SCIENTIA MAGNA An international journal

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2 Vol. 4, No. 3, 2008 ISSN SCIENTIA MAGNA An international journal Edited by Department of Mathematics Northwest University Xi an, Shaanxi, P.R.China

3 Scientia Magna is published annually in pages per volume and 1,000 copies. It is also available in microfilm format and can be ordered (online too) from: Books on Demand ProQuest Information & Learning 300 North Zeeb Road P.O. Box 1346 Ann Arbor, Michigan , USA Tel.: (Customer Service) URL: Scientia Magna is a referred journal: reviewed, indexed, cited by the following journals: "Zentralblatt Für Mathematik" (Germany), "Referativnyi Zhurnal" and "Matematika" (Academia Nauk, Russia), "Mathematical Reviews" (USA), "Computing Review" (USA), Institute for Scientific Information (PA, USA), "Library of Congress Subject Headings" (USA). Printed in the United States of America Price: US$ i

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5 Contributing to Scientia Magna Authors of papers in science (mathematics, physics, philosophy, psychology, sociology, linguistics) should submit manuscripts, by , to the Editor-in-Chief: Prof. Wenpeng Zhang Department of Mathematics Northwest University Xi an, Shaanxi, P.R.China Or anyone of the members of Editorial Board: Dr. W. B. Vasantha Kandasamy, Department of Mathematics, Indian Institute of Technology, IIT Madras, Chennai , Tamil Nadu, India. Dr. Larissa Borissova and Dmitri Rabounski, Sirenevi boulevard , Moscow , Russia. Dr. Huaning Liu, Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China. hnliu@nwu.edu.cn Prof. Yuan Yi, Research Center for Basic Science, Xi an Jiaotong University, Xi an, Shaanxi, P.R.China. yuanyi@mail.xjtu.edu.cn Dr. Zhefeng Xu, Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China. zfxu@nwu.edu.cn; zhefengxu@hotmail.com Dr. Zuoren Wang, School of Statistics, Xi an University of Finance and Economics, Xi an, Shaanxi, P.R.China. wangzuoren@163.com Dr. Yanrong Xue, Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China. xueyanrong1203@163.com iii

6 Contents F. Li : On the Smarandache additive sequence 1 M. Turgut and S. Yilmaz : Contributions to classical differential geometry of the curves in E 3 5 K. Wang : Almost everywhere stability of the third-ordered nonlinear system 10 H. Liu : On the indexes of beauty 14 Y. Liu : On the mean value of a new arithmetical function 16 Q. Wang and D. Zhang : A new decision making method for fuzzy multiple attribute decision making problem 19 J. Fu and Y. Wang : A new Smarandache function and its elementary properties 26 Y. Zhao : Some divisibility of Smarandache numerical carpet 29 Y. Yu and L. Cai : On the Smarandache prime part sequences and its two conjectures 33 S. Jin and P. Tang : On the continued fractions and Dirichlet L-functions 36 C. Liu and W. He : Research on the connectivity of public transportation network based on set method 41 Y. Xue : On the property of the Smarandache-Riemann zeta sequence 47 B. Chen and C. Shi : The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 50 M. Liang and L. Dai : An iterative method for (R,S)-symmetric solution of matrix equation AXB = C 60 L. Cai : On an equation of the Smarandache function 71 C. Wang : On the Smarandache divisibility theorem 74 M. Fang : Some properties of the permutation sequence 77 X. Qi : One problem related to the Smarandache quotients 79 A. A. Majumdar : The Smarandache bisymmetric determinant natural sequence 81 X. Li : On the k-power complements function of n! 86 A. A. Majumdar : An alternative approach to the LP problem with equality constraints 89 A. A. Majumdar : On some pseudo Smarandache function related triangles 95 W. Qin : One problem related to the Smarandache function 106 N. Dung : Notes on mssc-images of relatively compact metric spaces 109 iv

7 Scientia Magna Vol. 4 (2008), No. 3, 1-4 On the Smarandache additive sequence Fanbei Li School of Mathematics and Statistics, Inner Mongolia Finance and Economics College Hohhot , Inner Mongolia, P.R.China Abstract In this paper, we use the elementary method to study the properties of the Smarandache additive sequence, and give several interesting identities for some infinite series involving the Smarandache additive sequence. Keywords Smarandache additive sequence, infinite series, convergent properties. 1. Introduction and results For any positive integer m 2, let 1 < d 1 < d 2 < < d m are m positive integers, the Smarandache multiplicative sequence A m is defined as: If d 1, d 2,, d m are the first m terms of the sequence A m, then d k > d k 1, for k m + 1, is the smallest number equal to d α1 1 d α2 2 dαm m, where α i 1 for all i = 1, 2,, m. For example, the Smarandache multiplicative sequence A 4 ( generated by digits 2, 3, 5, 7 ) is: 2, 3, 5, 7, 210, 420, 630, 840, 1050, 1260, 1470, 1680, 1890, 2100,. In the book Sequences of Numbers Involved Unsolved Problems, Professor F.Smarandache introduced many sequences, functions and unsolved problems, one of them is the Smarandache multiplicative sequence, and he also asked us to study the properties of this sequence. About this problem, Ling Li [3] had studied it, and proved the following conclusion: For any positive integer m 2, let 1 < d 1 < d 2 < < d m are m positive integers, and A m denotes the Smarandache multiplicative sequence generated by d 1, d 2,, d m. Then for any real number s > 0, the infinite series is convergent, and n A m 1 n s 1 n s = m 1 m d s n A m i=1 i 1 + i=1 In this paper, we define another sequence called the Smarandache additive sequence as follows: Let 1 d 1 < d 2 < < d m are m positive integers, we define the Smarandache additive sequence D m as: If d 1, d 2,, d m are the first m terms of the sequence D m, then d k > d k 1, for k m + 1, is the smallest number equal to α 1 d 1 + α 2 d α m d m, 1. d s i

8 2 Fanbei Li No. 3 where α i 1 for all i = 1, 2,, m. For example, the Smarandache additive sequence D 2 ( generated by digits 3, 5 ) is: 3, 5, 8, 11, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,. About the properties of this sequence, it seems that none had studied it yet, at least we have not seen any related papers before. The problem is interesting, because it has the close relationship with the positive integer solutions of the indefinite equation d 1 x 1 + d 2 x 2 + d 3 x d m 1 x m 1 + d m x m = n. The main purpose of this paper is using the elementary method to study the convergent properties of some infinite series involving the Smarandache additive sequence, and get some interesting identities. For convenience, we use the symbol a m (n) denotes the n-th term of the Smarandache additive sequence D m. In this paper, we shall prove the following: Theorem. For any positive integer m 2, let 1 d 1 < d 2 < < d m are m positive integers, and D m denotes the Smarandache additive sequence generated by d 1, d 2,, d m. Then for any real number s 1, the infinite series 1 n s = 1 a s n D m n=1 m(n) (1) is divergent; For any real number s > 1, the series (1) is convergent, and 1 n s = 1 a s n D m n=1 m(n) = ζ(s) + a m(n)<m 1 a s m(n) n<m 1 n s, where ζ(s) is the Riemann zeta-function, and M is a computable positive integer. From this Theorem we may immediately deduce the following three corollaries: Corollary 1. Let D 2 be the Smarandache additive sequence generated by 2 and 3, then we have the identity 1 n 2 = 1 π2 a 2 = n D 2 n=1 2 (n) Corollary 2. Let D 2 be the Smarandache additive sequence generated by 4 and 6, then we have the identity 1 n 2 = 1 π2 a 2 = n D 2 n=1 2 (n) Corollary 3. Let D 3 be the Smarandache additive sequence generated by 3, 4 and 5, then we have the identity 1 n 2 = n D 3 n=1 a π2 = (n)

9 Vol. 4 On the Smarandache additive sequence 3 2. Proof of the theorem To prove our Theorem, we need a simple Lemma which stated as follows: Lemma. Let a and b are two positive integers with (a, b) = 1, then for any positive integer n > ab a b, the indefinite equation ax + by = n must has nonnegative integer solution. Proof. See Theorem 19 of Chapter one in reference [3]. Now we use this Lemma to prove our Theorem. First note that for any positive integer n > m, we have a m (n) = α 1 d 1 + α 2 d α m d m, where α i 1 for all i = 1, 2, 3,, m. From Lemma and its generalization we know that there exist a positive integer M such that for any integer n M, the indefinite equation d 1 x 1 + d 2 x 2 + d 3 x d m 1 x m 1 + d m x m = n must have positive integer solution. So from the definition of the sequence D m we know that for any integer n M, there must exist one term a m (k) in sequence D m such that n = a m (k). Therefore, 1 n s = 1 a s n D m n=1 m(n) = = a m(n) M n=1 1 a s m(n) + 1 n s = ζ(s) n<m n<m 1 n s + 1 n s + a m(n)<m a m(n)<m a m(n)<m 1 a s m(n) 1 a s m(n) 1 a s m(n), 1 where ζ(s) is the Riemann zeta-function, and the infinite series is convergent if s > 1; ns n=1 It is divergent if s 1. This completes the proof of Theorem. Now we prove our Corollary 1. Taking a = 2, b = 3, note that a b a b = = 1, so from Lemma we know that for all integer n 2, the indefinite equation 2x + 3y = n has nonnegative integer solution. Therefore, for all integer n 7, the indefinite equation 2x+3y = n has positive integer solution. So M = 7, from our Theorem we may immediately deduce that n=1 1 a 2 2 (n) = ζ(2) + a m(n)<7 1 a 2 m(n) n<7 = π n 2 n 6 = π , 1 n 2

10 4 Fanbei Li No. 3 where we have used the identity ζ(2) = π2. This proves Corollary 1. 6 To prove Corollary 2, note that the greatest common divisor of 4 and 6 is (4, 6) = 2, so from Corollary 1 we have the identity { 1 a 2 n=1 2 (n) = 1 ζ(2) } 1 n 2 n<7 = π Now we prove Corollary 3. Note that for all n 15, the indefinite equation 3x + 4y + 5z = n has positive integer solution. So we can take M = 15. Therefore, n=1 References 1 a 2 3 (n) = ζ(2) + a 3(n)<15 1 a 2 3 (n) n<15 1 n 2 = π [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] F. Smarandache, Sequences of Numbers Involved Unsolved Problems, Hexis, [3] Ling Li, On the Smarandache multiplicative sequence, Scientia Magna, 4(2008), No.2, [4] Luo Genghua, Introduction to number theory, Science Press, Beijing, [5] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, [6] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [7] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [8] I. Balacenoiu and V. Seleacu, History of the Smarandache function, Smarandache Notions Journal, 10(1999), No.1-2-3, [9] M. L. Perez, Florentin Smarandache Definitions, Solved and Unsolved Problems, Conjectures and Theorems in Number theory and Geometry, Chicago, Xiquan Publishing House, 2000.

11 Scientia Magna Vol. 4 (2008), No. 3, 5-9 Contributions to classical differential geometry of the curves in E 3 Melih Turgut and Suha Yilmaz Dokuz Eylul University, Buca Educational Faculty Department of Mathematics, 35160, Buca-Izmir,TURKEY melih.turgut@gmail.com Abstract In this paper, a system of differential equation whose solution gives the components on the Frenet axis in E 3 is established by means of Frenet equations. In view of some special solutions of mentioned system, position vector of rectifying curves, osculating curves with constant first curvature, normal curves and special cases are presented. characterization for inclined spherical curves is expressed. Keywords Euclidean space, Frenet frame. Moreover, a 1. Introduction It is safe to report that the many important results in the theory of the curves in E 3 were initiated by G. Monge; and G. Darboux pionnered the moving frame idea. Thereafter, F. Frenet defined his moving frame and his special equations which play important role in mechanics and kinematics as well as in differential geometry (for more details see [2]). In the Euclidean space E 3, it is well-known that to each unit speed curve with at least four continuous derivatives, one can associate three mutually orthogonal unit vector fields T, N and B are respectively, the tangent, the principal normal and the binormal vector fields[3]. At each of the curve, the planes spanned by {T, N}, {T, B} and {N, B} are known respectively as the osculating plane, the rectifying plane and the normal plane[5]. The curves ϕ : I R E 3 for which the position vector ϕ always lie in their rectifying plane, are simplicity called rectifying curves [1]. Similarly, the curves for which the position vector ϕ always lie in their osculating plane, are for simplicity called osculating curves and the curves for which the position vector always lie in their normal plane, are for simplicity called normal curves [4]. In this paper, position vectors of some special curves in E 3 are investigated. First, a system of differential equation is established. In view of some special solutions, characterizations of some special curves are presented.

12 6 Melih Turgut and Suha Yilmaz No Preliminaries To meet the requirements in the next sections, here, the basic elements of the theory of curves in the space E 3 are briefly presented. (A more complete elementary treatment can be found in [5].) The Euclidean 3-space E 3 provided with the standard flat metric given by, = dx dx dx 2 3, where (x 1, x 2, x 3 ) is a rectangular coordinate system of E 3. Recall that, the norm of an arbitrary vector a E 3 is given by a = a, a. ϕ is called an unit speed curve if velocity vector v of ϕ satisfies v = 1. For vectors v, w E 3 it is said to be orthogonal if and only if v, w = 0. Let ϑ = ϑ(s) be a curve in E 3. If the tangent vector field of this curve forms a constant angle with a constant vector field U, then this curve is called an inclined curve. The sphere of radius r > 0 and with center in the origin in the space E 3 is defined by S 2 = { p = (p 1, p 2, p 3 ) E 3 : p, p = r 2}. Denote by {T, N, B} the moving Frenet frame along the curve ϕ in the space E 3. For an arbitrary curve ϕ with first and second curvature, κ and τ in the space E 3, the following Frenet formulae are given in [3], [5] where T N B = 0 κ 0 T κ 0 τ N 0 τ 0 B, (1) T, T = N, N = B, B = 1, T, N = T, B = T, N = N, B = 0. Here, first and second curvature are defined by κ = κ(s) = T (s) and τ(s) = N, B. It is well-known that for a unit speed curve with non vanishing curvatures following proposition holds [3]. Proposition 2.1. Let ϕ = ϕ(s) be a regular curve with curvatures κ and τ, then; i. It lies on a sphere (S 2 ) if and only if [ τ κ + 1 τ ii. It is an inclined curve if and only if κ τ ( ) ] 1 = 0. (2) κ = constant. (3)

13 Vol. 4 Contributions to classical differential geometry of the curves in E Position vectors of some special curves in E 3 Let ϕ = ϕ(s) be a unit speed curve in E 3. We can write this curve respect to Frenet frame {T, N, B} as ϕ = ϕ(s) = m 1 T + m 2 N + m 3 B. (4) Differentiating (4) and considering Frenet equations, we have a system of differential equation as follow: m 1 m 2 κ 1 = 0 m 2 + m 1 κ m 3 τ = 0 m 3 + m 2 τ = 0. (5) This system s general solution have not been found. Due to this, we give some special values to the components. Case 3.1. m 1 = c 1 =constant. From (5) 1 we have second component And (5) 2 gives us third component as m 3 = 1 τ m 2 = 1 κ. (6) ( ) 1 κ + c 1 κ τ. (7) m 2 and m 3 satisfie (5) 3, if we rewrite it, we have following differential equation: [ ( ) ] 1 1 ( κ ) τ + c 1 = 0. (8) τ κ τ κ Considering (8) and Proposition 2.1., suffice it to say that following theorem holds: Theorem 3.1. Let ϕ = ϕ(s) be a unit speed curve and first component of position vector on Frenet axis be constant in E 3 : i. If first component is zero, then ϕ is a spherical curve (lies on S 2 ). ii. If ϕ is an inclined curve, then ϕ lies on S 2. Therefore, ϕ is an inclined spherical curve. iii. All normal curves in the space E 3 are spherical curves. iv. Position vector all of normal curves in E 3, can be written as follow: ϕ = 1 κ N 1 ( ) 1 B. (9) τ κ Here, it is not difficult to see that statement ii is a characterization for inclined spherical curves in E 3. Case 3.2. m 2 = c 2 = constant 0. From equation (5) 1, we obtain From (5) 2, we write c 2 = m 1 1. (10) κ m 3 = m 1 κ τ. (11)

14 8 Melih Turgut and Suha Yilmaz No. 3 Now, let us suppose ϕ is an inclined curve. In this case, If (10) and (11) are used in (5) 3, we have a differential equation as follow: From solution of (12), we have the components and m 1( κ τ + τ κ ) τ = 0. (12) κ m 1 = m 2 = τ 2 s κ 2 + l, (13) + τ 2 κ2 κ 2 + τ 2 (14) m 3 = κ τ ( τ 2 s κ 2 + l), (15) + τ 2 where l R. Corollary 3.2. Let ϕ = ϕ(s) be a unit speed curve and second component of position vector on Frenet axis be constant in E 3. Then position vector of ϕ can be formed by the equations (13), (14) and (15). Case m 2 = 0. From (5), we have, respectively, and m 1 = s + l (16) m 3 = κ τ (s + l) = constant, (17) where l R. Considering (16) and (17), we give following theorem. Theorem 3.3. Let ϕ = ϕ(s) be a unit speed rectifying curve in E 3. Then; i. There is a relation among curvatures as κ (s + l) = constant. τ ii. Position vector of ϕ can be written as follow: ϕ = ϕ(s) = (s + l)t + κ τ (s + l) B, (18) where l R. Remark 3.4. The case m 3 = constant 0 is similar to case 3.2. Case 3.3. m 3 = 0. Thus τ = 0. And here, let us suppose κ = constant. Using these in the system of differential equation, we have second order homogenous differential equation with constant coefficient as follow: Solution of (19) is elementary. Then, we write the components m 2 + m 2 κ 2 + κ = 0. (19) m 1 = A 2 sin κs A 1 cos κs (20)

15 Vol. 4 Contributions to classical differential geometry of the curves in E 3 9 and m 2 = A 1 cos κs + A 2 sin κs + 1 κ, (21) where A 1 and A 2 are real numbers. Corollary 3.5. Let ϕ = ϕ(s) be a unit speed osculating curve with constant first curvature in E 3. The position vector of ϕ can be composed with the equations (20) and (21). References [1] B. Y. Chen, When does the position vector of a space curve always lie in its rectifying plane?, Amer. Math. Mounthly, 110 (2003), No.2, [2] C. Boyer, A History of Mathematics, New York, Wiley, [3] H. H. Hacisalihoglu, Differential Geometry, Ankara University Faculty of Science Press, [4] K. Ilarslan, Space-like Normal Curves in Minkowski Space E 3 1, Turk J. Math., 29(2005), [5] R. S. Milman and G. D. Parker, Elements of Differential Geometry, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1977.

16 Scientia Magna Vol. 4 (2008), No. 3, Almost everywhere stability of the third-ordered nonlinear system Kaiming Wang School of Science, Chang an University, Xi an, Shaanxi, P.R.China kaimingwang@yahoo.com.cn Abstract In order to study the stability of the third-ordered nonlinear system, density function is constructed, and an inequality condition is obtained. When the considered system satisfies the inequality for almost all initial value, its equilibrium is asymptotically stable almost everywhere. Keywords Amost everywhere stability, density function, nonlinear system. 1. Introduction Stability is a very important factor in the research of high-ordered nonlinear systems. The techniques currently available in the literature are mostly based on Lyapunov s direct method [1], contraction mapping method [2], nonlinear measure method [3,4], nonlinear spectral radius method [5] and so on. All these methods make the research of stable property everywhere. And if the equilibrium of the considered system is asymptotically stable, there exists a neighborhood of the equilibrium which is called attraction region. In fact, many real nonlinear systems do not need its equilibrium acting as an attracting actor attracting any points of its neighborhood, and it only needs attract enough many points in this neighborhood. That is, the equilibrium is asymptotically stable almost everywhere (the equilibrium attract all points of this neighborhood except a set with zero Lebesgue measure). In 2001, Anders Ranzter [6] proposed the concept of a dual to Lyapunov s stability theorem, and he called the stability derived by this way as stability almost everywhere. In control theory, most control systems are in high order, such as robotic multi-fingered grasping system [1], neural networks and so on, and up to their research background, it s unnecessary or impossible for the equilibrium to attract every points in its neighborhood even it s very small. So it s valuable to study the almost everywhere stability of high-ordered systems. In this paper, the almost everywhere stability of the third-ordered system is studied. Density functions are constructed as a tool to analyze the stability of the given system. For convenience, the following notations were used in this paper: (1) ρ = gradρ = ( ρ x 1 ρ x 2 ρ x n ), ρ : R n R; (2) F = divf = F1 x 1 + F2 x Fn x n, F : R n R n ; (3) represents inner product.

17 Vol. 4 Almost Everywhere Stability of the Third-ordered nonlinear system Main result and its proof Consider a third-ordered nonlinear system or its equivalent system x + ψ(x, x )x + f(x, x ) = 0, f(0, 0) = 0, (1) x y z = y z ψ(x, y)z f(x, y) It is assumed that solutions of (1) exist and are unique. = F (x, y, z). (2) Theoretically, this is a very useful nonlinear system since (1) is a rather general thirdordered nonlinear differential equation. For example, many third-ordered differential equations which have been discussed in [8] are special cases of (1). However, most available research all focused on the study of everywhere stability and it is not necessary or impossible for many real nonlinear equations. In this section, the almost everywhere stability of third-ordered system is studied. Density functions are constructed as a tool to analyze the stability of the given system, and the condition was obtained that the given nonlinear system is asymptotically stable almost everywhere. That is, we will prove the following main result: Theorem. If there exists a positive real number α, such that the following inequalities (1) xf(x, y) > 0, yf(x, y) > 0; (2) ψ(x, y) > 0; y (3) 0 (f(x, 0) + yψ x(x, v)v + yf x (x, v) + zψ(x, y)) dv+ +α 1 ψ(x, y) ( x 0 f(u, 0)du + y (vψ(x, v) + f(x, v))dv)) 0 < ψ(x, y)z 2 (1 1 2 α 1 ) hold for almost all x, then the equilibrium of the nonlinear system (2) is asymptotically stable almost everywhere. First, we need a basic Lemma that is necessary in proof of Theorem. Lemma. Given the equation ẋ(t) = f(x(t)), where f C 1 (R n, R n ) and f(0) = 0, suppose there exists a non-negative scalar function ρ C 1 (R n \ {0}, R) such that ρ(x)f(x)/ x is integrable on {x R n : x 1} and [ (fρ)](x) > 0 for almost all x. Then, for almost all initial states x(0) the trajectory x(t) exists for t (0, ) and tends to zero as t. Moreover, if the equilibrium x = 0 is stable, then the conclusion remains valid even if ρ takes negative values. Proof. ( See reference [6] ). By this Lemma, now we shall complete the proof of Theorem. The proof of Theorem. Let x = (x, y, z) and V ( x) = x 0 f(u, 0)du + y 0 vψ(x, v)dv + y 0 f(x, v)dv z2,

18 12 Kaiming Wang No. 3 then V ( x) 0 and when x 0 we have V ( x) 0. That is, V ( x) is positive definite. Let ρ = V α, then ρ C 1 (R 3 \{0}, R) and ρ( x)f ( x)/ x is integrable on { x R 3 : x 1}. For divf = ψ(x, y) and [ (F ρ)]( x) = V α 1 ( x)(v F α V F )( x), thus the condition [ (F ρ)]( x) > 0 in Lemma will be satisfied if and only if (V F α V F )( x) > 0. From and V F = V x x + V y y + V z z = f(x, 0)y + y y ψ x (x, v)vdv + y y 0 0 f x (x, v)dv ψ(x, y)z 2 + ψ(x, y)yz, α 1 V F = α 1 ψ(x, y) α 1 ψ(x, y) x 0 y then by Lemma, we know that if the inequality y 0 0 f(u, 0)du α 1 ψ(x, y) y f(x, v)dv 1 2 α 1 z 2 ψ(x, y), 0 vψ(x, v)dv (f(x, 0) + yψ x (x, v)v + yf x (x, v) + zψ(x, y)) dv ( x y ) +α 1 ψ(x, y) f(u, 0)du + (vψ(x, v) + f(x, v))dv) < ψ(x, y)z 2 (1 1 2 α 1 ) 0 holds, the equilibrium of the considered system is asymptotically stable almost everywhere. This completes the proof of Theorem. Simplifying the condition of the inequality in Theorem, we may immediately derive the following two corollaries. Corollary 1. If there exists a positive real number α > 1, such that two functions f(x, y) 2 and ψ(x, y) satisfy the inequalities (1) xf(x, y) > 0, yf(x, y) > 0; (2) ψ(x, y) > 0; y 0 (f(x, 0) + yψ x(x, v)v + yf x (x, v) + ψ(x, y)z)dv < 0; (3) x (4) 0 f(u, 0)du + y 0 (vψ(x, v) + f(x, v))dv < z2 (α 1 2 ) for almost all x, then the equilibrium of the nonlinear system (2) is asymptotically stable almost everywhere. Corollary 2. If there exists a positive real number α > 1,such that two functions f(x, y) 2 and ψ(x, y) satisfy the inequalities (1) xf(x, y) > 0, yf(x, y) > 0; (2) ψ(x, y) > 0; 0

19 Vol. 4 Almost Everywhere Stability of the Third-ordered nonlinear system 13 (3) yf(x, 0) + yzψ(x, y) + y 2 (ψ x (x, ξ)ξ + f x (x, η)) < 0, ξ, η [0, y]; (4) xf(s, 0) + y(hψ(x, h) + f(h, 0)) < z 2 (α 1 ), s [0, x], h [0, y] 2 for almost all x, then the equilibrium of the nonlinear system (2) is asymptotically stable almost everywhere. Since the system (1) and system (2) are equivalent systems, and the equilibrium x = 0 of system (1) answers to the equilibrium x = 0 of system (2), their stability properties are also equivalent. Thus the equilibrium of the third-ordered nonlinear system (1) is almost asymptotically stable when the inequality condition of Theorem is satisfied. References [1] C. H. Xiong, Y. F. Li, Y. L. Xiong, H. Ding, On the dynamic stability of grasping, International Journal of I Robotics Research, 18(1999), [2] Kelly D.G, Stability in contractive nonlinear neural networks, IEEE Trans Biomed Eng., 37(1990), [3] Hong Qiao, J. G. Peng and Z. B. Xu, Nonlinear Measure: A new approach to exponential stability analysis for Hopfield-type neural networks, IEEE Transactions on Neural Networks 12(2001), [4] K. M. Wang, S. Y. Jia, X. L. Hu, Dynamic stability of robotic multi-fingered grasping based on nonlinear measure, Journal of Chang an University, Natural Science Edition, 27(2007), [5] K. L. Mak, J. G. Peng, Z. B. Xu, K. F. C. Yiu, A new stability criterion for discrete-time neural networks, Nonlinear spectral radius, Chaos Solitons & Fractals, 31(2007), [6] Rantzer. A, A dual to Lyapunov s tability theorem, Systems & Control Letters, 42(2001), [7] M. O. Omeike, Further results on global stability of third-order nonlinear differential equations, Nonlinear Analysis, 67(2007), [8] R. Reissig, G. Sansone and R. Conti, Nonlinear Differential Equations of Higher Order, Leyden, Noordhoff Inter. Pub., [9] J. O. C. Ezeilo, On the stability of solutions of certain differential equations of the third order, Quart. J. Math., Oxford Ser., 11(1960),

20 Scientia Magna Vol. 4(2008), No. 3, On the indexes of beauty 1 Hongyan Liu Department of Applied Mathematics, Xi an University of Technology, Xi an, P.R.China Abstract For any fixed positive integer m, if there exists a positive integer n such that n = m d(n), then m is called an index of beauty, where d(n) is the Dirichlet divisor function. In this paper, we shall prove that there exist infinite positive integers such that each of them is not an index of beauty. Keywords Dirichlet divisor function, index of beauty, elementary method. 1. Introduction and result For any positive integer n, the famous Dirichlet divisor function d(n) is defined as the number of all distinct divisors of n. If n = p α1 1 pα2 2 pα k k be the prime power factorization of n, then from the definition and properties of d(n) we can easily get d(n) = (α 1 + 1) (α 2 + 1) (α k + 1). (1) About the deeply properties of d(n), many people had studied it, and obtained a series results, see references [2], [3] and [4]. In reference [5], Murthy introduced an index of beauty involving function d(n) as follows: For any positive integer m, if there exists a positive integer n such that m = n d(n), then m is called an index of beauty. At the same time, he also proposed the following conjecture: Conjecture. Every positive integer is an index of beauty. Maohua Le [6] gave a counter-example, and proved that 64 is not an index of beauty. We think that the conclusion in [6] can be generalization. This paper as a note of [6], we shall use the elementary method to prove the following general conclusion: Theorem. There exists a set A including infinite positive integers such that each of n A is not an index of beauty. 2. Proof of the theorem In this section, we shall prove our Theorem directly. For any prime p 5, if we taking m = p p 1, then we can prove that m is not an index of beauty. In fact, if m is an index of beauty, then there exists a positive integer n such that n = p p 1 d(n). From this identity we 1 This work is supported by the Shaanxi Provincial Education Department Foundation 08JK398.

21 Vol. 4 On the indexes of beauty 15 can deduce that p p 1 have the identity n. Let n = p α b and (p, b) = 1, then from n = p p 1 d(n) and (1) we p α p+1 b = (α + 1) d(b). (2) From (2) we may immediately deduce that α p 1. If α = p 1, then (2) become b = p d(b), this contradiction with (b, p) = 1. If α = p, then (2) become p b = (p + 1) d(b), or ( b = ) d(b). (3) p It is clear that (3) is not possible if prime p 5. In fact if b = 1, 2, 3, 4, 5, 6, 7, then it is easily to check that (3) does not hold. If b 8, then from (1) and the properties of d(b) we can deduce that b 3 ( 2 d(b) > ) d(b). So (3) does not also hold. So if α = p, then (2) is p not possible. If α p + 1, let α p = k 1, the (2) become p k+1 b = (p + k + 1) d(b). (4) Note that b d(b) for all positive integer b, so formula (4) is not possible, since p k+1 > p+k +1 for all prime p 5 and k 1. Since there are infinite prime p 5, and each p p 1 is not an index of beauty, so there exists a set A including infinite positive integers such that each of n A is not an index of beauty. This completes the proof of Theorem. References [1] F.Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] Pan Chengdong and Pan Chengbiao, The elementary proof of the prime theorem (in Chinese), Shanghai Science and Technology Press, Shanghai, [3] Tom M. Apostol, Introduction to Analytical Number Theory, Spring-Verlag, New York, [4] Aleksandar Ivić, The Riemann Zeta-Function, Dover Publications, New York, [5] A. Murthy, Some more conjectures on primes and divisors, Smarandache Notions Journal, 12(2001), [6] Mohua Le, A conjecture concerning indexes of beauty, Smarandache Notions Journal, 14(2004), [7] M. L. Perez, Florentin Smarandache, definitions, solved and unsolved problems, conjectures and theorems in number theory and geometry, Xiquan Publishing House, [8] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, [9] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [10] Luo Genghua, Introduction to number theory, Science Press, Beijing, [11] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, Oxford University Press, Oxford, 1937.

22 Scientia Magna Vol. 4 (2008), No. 3, On the mean value of a new arithmetical function Yanyan Liu Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any positive integer n, let φ(n) denotes the Euler function. That is, φ(n) denotes the number of all positive integers not exceeding n which are relatively prime to n. g(n) denotes the smallest positive integer m such that n φ(n) m. The main purpose of this paper is using the elementary method ( to prove that for any positive integer group k ) k (m 1, m 2,, m k ), we have the inequality g m i g(m i), and also for any positive integer ( k, there exist infinite group positive integers (m 1, m 2,, m k ) satisfying the equation k ) k g m i = g(m i). i=1 i=1 Keywords Euler function φ(n), three prime theorem, elementary method. i=1 i=1 1. Introduction and results For any positive integer n, the famous Euler function φ(n) is defined as the number of all positive integers not exceeding n which are relatively prime to n. Now we define a new arithmetical function g(n) as follows: g(n) denotes the smallest positive integer m such that n φ(n) m. That is, g(n) = min{m : n φ(n) m}. For example, the fist few values of g(n) are: g(1) = 1, g(2) = 2, g(3) = 3, g(4) = 2, g(5) = 5, g(6) = 3, g(7) = 7, g(8) = 2, g(9) = 3, g(10) = 5, g(11) = 11, g(12) = 3, g(13) = 13, g(14) = 7,. About the elementary properties of g(n), it seems that none had studied it before. From the definition of g(n) we can easily deduce that for any prime p, g(p α ) = p and P (n) g(n) p n p, where P (n) denotes the largest prime divisor of n. If n > 1 and n 2 α, then g(n) must be an odd number. g(n) is also a square-free number. The main purpose of this paper is to study the elementary properties of g(n), and prove the following two conclusions: Theorem 1. For any positive integer k and any positive integer group (m 1, m 2,, m k ), we have the inequality ( k ) k g m i g(m i ). i=1 Theorem 2. For any positive integer k 3, there exist infinite group positive integers i=1

23 Vol. 4 On the mean value of a new arithmetical function 17 (m 1, m 2,, m k ) such that the equation ( k ) k g m i = g(m i ). i=1 i=1 Whether there exist infinite group positive integers (m 1, m 2 ) satisfying the equation g(m 1 + m 2 ) = g(m 1 ) + g(m 2 ) (1) is an unsolved problem. If m 1 > 1 and m 2 > 1 satisfying the equation (1), then one of m 1 and m 2 must be a power of 2. It is an interesting problem to study the mean value properties of g(n). number x > 1, we conjecture that where C is a computable constant. g(n) C x 2, n x For any real 2. Proof of the theorems In this section, we shall use the elementary method to prove our Theorems. First we prove Theorem 1. For any positive integer m, since g(m) = g(m). So Theorem 1 holds if k = 1. Now we assume k = 2. For any positive integers m and n, if (m, n) = 1, then we have φ(mn) = φ(m)φ(n). This time let g(mn) = t, g(m) = u, g(n) = v. From the definition of g(n) we have mn φ(mn) t, m φ(m) u, n φ(n) v, so we have mn φ(m) φ(n) uv = φ(mn) uv. By the definition of t we have t uv. Therefore, g(mn) g(m) g(n). If (m, n) > 1, without loss of generality we can assume that m = p α and n = p β, where p be a prime, α and β are two positive integers, then from the definition of g(n) we have g(p α p β ) = p, g(p α ) = p, g(p β ) = p. So g(p α p β ) g(p α ) g(p β ). From the above we know that for any positive integers m and n, we have the inequality g(mn) g(m) g(n). (2) Now for any positive integer k > 2 and any positive integer group (m 1, m 2,, m k ), from (2) we have ( k ) ( g m i = g m k i=1 This proves theorem 1. k 1 i=1 m i ) g ( k 1 i=1 m i ) g(m k ) k g(m i ). To prove Theorem 2, we need the famous Vinogradov s Three Prime Theorem, which was stated as follows: i=1

24 18 Yanyan Liu No. 3 Lemma 1. There exists a sufficiently large constant K > 0 such that each odd integer n > K can be written as a sum of three odd primes. That is, n = p 1 + p 2 + p 3, where p i (i = 1, 2, 3) are three odd primes. Proof. (See reference [1]). Lemma 2. Let k 3 be an odd integer, then any sufficiently large odd integer n can be written as a sum of k odd primes n = p 1 + p p k. Proof. (See reference [2]). Now we use these two Lemmas to prove our Theorem 2. If k = 3, then from Lemma 1 we know that for any prime p large enough, it can be written as a sum of three primes: p = p 1 + p 2 + p 3. Now for any positive integer k 4 any prime p large enough, if k be an odd number, then by the definition of g(n) we know that g(p) = p. From Lemma 2 we may immediately get p = g(p) = g(p 1 + p p k ) = p 1 + p p k = g(p 1 ) + + g(p k ). If k 4 be an even number, then k 1 3 be an odd number. So if prime p large enough, then p 2 can be written as a sum of k 1 primes: This implies or p 2 = p 1 + p p k 1. p = 2 + p 1 + p p k 1 p = g(p) = g(2 + p 1 + p p k 1 ) = 2 + p 1 + p p k 1 = g(2) + g(p 1 ) + + g(p k 1 ). Since there are infinite primes p, so there exist infinite group positive integers (m 1, m 2, m k ) satisyfying the equation ( k ) k g m i = g(m i ). i=1 This completes the proof of Theorem 2. i=1 References [1] Pan Chengdong and Pan Chengbiao, Goldbach Conjecture, Science Press, Beijing, [2] Lu Yaming, On the solution of an equation involing the Smarandache function, Scientia Magna, 2(2006), No. 1, [3] Pan Chengdong and Pan Chengbiao, The elementary proof of the prime theorem (in Chinese), Shanghai Science and Technology Press, Shanghai, [4] Tom M. Apostol, Introduction to analytical number theory, Spring-Verlag, New York, 1976.

25 Scientia Magna Vol. 4(2008), No. 3, A new decision making method for fuzzy multiple attribute decision making problem 1 Qiuping Wang and Daohong Zhang Department of Applied Mathematics, Xi an University of Technology, Xi an, Shaanxi , P.R.China School of Business Management, Xi an University of Technology, Xi an, Shaanxi , P.R.China wqp566@yahoo.com.cn Abstract It is proved that product of interval number and triangular fuzzy number is a trapezoidal fuzzy number. This paper transformed a multiple attribute decision making problem with attribute weight of interval number and attribute value of triangular fuzzy number into trapezoidal fuzzy number multiple attribute decision making problem, gave TOPSIS to trapezoidal number multiple attribute decision making problem, and analyzed a practical problem by this method. Keywords Interval number, triangular fuzzy number, trapezoidal fuzzy number, multiple attribute decision making, TOPSIS. 1. Introduction The multiple decision making (MADM) problems mainly solve to rank alternatives with multiple attributes among finite alternatives. It is an extremely active topic in discipline research of modern decision science, systems engineering, management science and operations research, etc. Its theory and method are applied to many domains: engineering design, economy, management and military affairs, etc. Reference [1],[2],[3] studied MADM problems that attribute weight information is interval number, attribute value is crisp number from various angles. Reference [4]-[7] studied MADM problems that attribute weight information is interval number, attribute value is also interval number from various angles. At present research results about methods for MADM problems with attribute weight of interval number and attribute value of triangular fuzzy number are few. This paper focuses on this kind of MADM problems mainly. This paper will prove that product of interval number and the triangular fuzzy number is a trapezoidal fuzzy number and transform a multiple attribute decision making problem with attribute weight of interval number and attribute value of and triangular fuzzy number into trapezoidal fuzzy number multiple attribute decision making problem, then gives TOPSIS to 1 This work is supported by scientific research plan foundation of Xi an University of Technology(210406)

26 20 Qiuping Wang and Daohong Zhang No. 3 trapezoidal number multiple attribute decision making problem. Finally, an example is used to illustrate the proposed method. 2. Presentation of the problem Definition 1. Let a be interval number, where a = [a, a + ] = {x a x a +, a a + }. If 0 a a +, then a is called a positive interval number. Definition 2. Fuzzy number p is called a triangular fuzzy number if its membership function µ p (x) : R [0, 1] is described as x m l l, x [l, m] m l x µ p (x) = m u u, x [m, u] m u 0, otherwise where x R, l m u, l, u stand for the lower and upper bounds of the support of p respectively, and m for the most possible value of p. The triangular fuzzy number p can be denoted by (l, m, u). When l = m = u, p is a real number. Definition 3. Fuzzy number t is called a trapezoidal fuzzy number if its membership function µ t (x) : R [0, 1] is described as µ t (x) = 0, x < a x a b a, a x < b 1, b x < c d x d c, c x < d 0, d x where a b c d, a and d stand for the lower and upper bounds of the support of t respectively, and closed interval [b, c] for the most possible value of t. The trapezoidal fuzzy number t can be denoted by (a, b, c, d). When a = b, c = d, t becomes a ordinary interval number. When b = c, t turns into triangular fuzzy number. Interval number and triangular fuzzy number are special cases of trapezoidal fuzzy number. When a = b = c = d, it is a ordinary real number. The λ-cut of a trapezoidal fuzzy number t = (a, b, c, d) is t λ = [a + λ(b a), d + λ(c d)], λ (0, 1]. Definition 4.[8] Let vector T = ( t 1, t 2,, t m ) and Ṽ = (ṽ 1, ṽ 2,, ṽ m ), where trapezoidal fuzzy number t i = ( t 1i, t 2i, t 3i, t 4i ), ṽ i = (ṽ 1i, ṽ 2i, ṽ 3i, ṽ 4i ), i = 1, 2,, m. The distance between T and Ṽ is defined as D( T, Ṽ ) = T Ṽ = 1 m [(t 1i v 1i ) (t 2i v 2i ) 2 + (t 3i v 3i ) 2 + (t 4i v 4i ) 2 ]. i=1

27 Vol. 4 A new decision making method for fuzzy multiple attribute decision making problem 21 We will use product of interval number and triangular fuzzy number in our discussion, for this reason we prove the theorem as follows. Theorem. Let α = [α 1, α 2 ] be arbitrary positive interval number, p = (l, m, n) be arbitrary triangular fuzzy number, then the product of α and p is α p = [α 1, α 2 ] (l, m, u) = (α 1 l, α 1 m, α 2 m, α 2 u). Namely the product of positive interval number and triangular fuzzy number is trapezoidal fuzzy number. Proof. For λ (0, 1], the λ-cut of p is According to extension principle we have α p = = p λ = [l + λ(m l), u + λ(m u)]. 0<λ 1 0<λ 1 λ{[α 1, α 2 ][l + λ(m l), u + λ(m u)]} λ[α 1 l + λ(α 1 m α 1 l), α 1 u + λ(α 2 m α 2 u)] = (α 1 l, α 1 m, α 2 m, α 2 u). 3. Principle and step This paper considers the fuzzy MADM problems that the alternatives are X 1, X 2,, X n and attributes are G 1, G 2,, G m. The fuzzy MADM problems with incomplete weight information, w j denotes the weight of attribute G j, w j [α j, β j ], where, 0 α j β j 1, j = 1, 2,, m, w 1 + w w m = 1, is known to us. Triangular fuzzy number (l ij, m ij, u ij ) is the value for alternative X i with respect to attribute G j, i = 1, 2,, n, j = 1, 2,, m. Principle and step for the fuzzy MADM problems that weight of attribute is interval number and value of attribute is triangular fuzzy number are as follows: Step 1. Construct triangular fuzzy number decision-making matrix à = (x ij) n m. Where triangular fuzzy number x ij = (l ij, m ij, u ij ) is the value for alternative X i with respect to attribute G j, i = 1, 2,, n, j = 1, 2,, m. Step 2. Normalize decision-making matrix Ã. Generally speaking, the attribute values have different physical dimension from each other, the decision matrix needs to be normalized so as to transform the various attribute values into comparable values. Therefore, we can obtain the normalized fuzzy decision matrix denoted by A = (x ij ) n m according to formulas in reference [9]. Let x ij = (l ij, m ij, u ij ), then x ij = ( l ij u max j, m ij u max j, u ij u max j ), j J 1, x ij = ( lmin j, lmin j, lmin j ), j J 2, u ij m ij l ij where u max j = max u ij, if j J 1, lj min = min l ij, if j J 2, where J 1 and J 2 are the set of benefit i i criteria and cost criteria, respectively.

28 22 Qiuping Wang and Daohong Zhang No. 3 Step 3. Construct the weighted normalized fuzzy decision matrix. w 1 x 11 w 2 x 12 w m x 1m ỹ 11 ỹ 12 ỹ 1m w 1 x 21 w 2 x 22 w m x 2m ỹ 21 ỹ 22 ỹ 2m Ỹ = W A = = w 1 x n1 w 2 x n2 w m x nm ỹ n1 ỹ n2 ỹ nm where ỹ ij = w j x ij = [α j, β j ] (l ij, m ij, u ij ) = (α j l ij, α j m ij, β j m ij, β j u ij ) = (a ij, b ij, c ij, d ij ), i = 1, 2,, n, j = 1, 2,, m. Step 4. Determine the fuzzy positive-ideal solution X + and fuzzy negative-ideal solution X, respectively, and where X + = (y + 1, y+ 2,, y+ m), X = (y 1, y 2,, y m) y + j = ( max a ij, max b ij, max c ij, max d ij) = (a + j, 1 i n 1 i n 1 i n 1 i n b+ j, c+ j, d+ j ), j = 1, 2,, m, y j = ( min a ij, min b ij, min c ij, min d ij) = (a j, 1 i n 1 i n 1 i n 1 i n b j, c j, d j ), j = 1, 2,, m. Step 5. Calculate the distance of each alternative from fuzzy positive-ideal solution X + and fuzzy negative-ideal solution X, respectively. The distance of each alternative from X + and X can be calculated as D + i = D(X i, X + ) = 1 m [(a ij a + j 4 )2 + (b ij b + j )2 + (c ij c + j )2 + (d ij d + j )2 ], j=1 D i = D(X i, X ) = 1 m [(a ij a j 4 )2 + (b ij b j )2 + (c ij c j )2 + (d ij d j )2 ], j=1 i = 1, 2,, n. Step 6. Calculate the relative closeness coefficient of each alternative to the ideal solution. C i = D i D i + D + i, i = 1, 2,, n. Step 7. Determine the ranking order of all alternatives according to the closeness coefficient in the descending order and select the best one.

29 Vol. 4 A new decision making method for fuzzy multiple attribute decision making problem Numerical example This example is taken from [10], one of the Middle East country intends to purchase a certain number jet fighter from US, The American Pentagon s officials have provided information relative to 4 kinds of jet fighter types which are permitted to sell. The Middle East country sends out the expert group to carry on the detailed inspection to these 4 kinds of jet fighters, Inspection results are shown in Table 1. Which kind of jet fighter should they purchase to make total utility value of decision making biggest? Table 1: Expert group inspection result to 4 kinds of jet fighters Type Maximum Cruising Peak load Price Reliability Maintainability speed (Ma) radius (mile) (pond) (/$ 10 6 ) X High Very high X Low General X High High X General General weight w j of attribute G j cannot determine completely (j = 1, 2,, 6), but according to the expert analysis we know w 1 [0.15, 0.20], w 2 [0.1, 0.15], w 3 [0.1, 0.15], w 4 [0.1, 0.15], w 5 [0.15, 0.20], w 6 [0.2, 0.25]. Step 1. Correspond to normalized fuzzy decision-making matrix as follows: (0.8, 0.8, 0.8) (0.56, 0.56, 0.56) (0.95, 0.95, 0.95) (0.82, 0.82, 0.82) (0.6, 0.8, 1) (0.7, 0.9, 1) (1, 1, 1) (1, 1, 1) (0.86, 0.86, 0.86) (0.69, 0.69, 0.69) (0, 0.2, 0.4) (0.3, 0.5, 0.7) (0.72, 0.72, 0.72) (0.74, 0.74, 0.74) (1, 1, 1) (1, 1, 1) (0.6, 0.8, 1) (0.6, 0.8, 1) (0.88, 0.88, 0.88) (0.67, 0.67, 0.67) (0.95, 0.95, 0.95) (0.9, 0.9, 0.9) (0.3, 0.5, 0.7) (0.3, 0.5, 0.7) Step 2. Construct the weighted normalized fuzzy decision matrix. (0.12, 0.12, 0.16, 0.16) (0.056, 0.056, 0.084, 0.084) (0.095, 0.095, , ) (0.15, 0.15, 0.2, 0.2) (0.1, 0.1, 0.15, 0.15) (0.086, 0.086, 0.129, 0.129) (0.108, 0.108, 0.144, 0.144) (0.074, 0.074, 0.111, 0.111) (0.1, 0.1, 0.15, 0.15) (0.132, 0.132, 0.176, 0.176) (0.067, 0.067, , ) (0.095, 0.095, , ) (0.082, 0.082, 0.123, 0.123) (0.09, 0.12, 0.16, 0.20) (0.14, 0.18, 0.225, 0.25) (0.069, 0.069, , ) (0, 0.03, 0.04, 0.08) (0.06, 0.1, 0.125, 0.175) (0.1, 0.1, 0.15, 0.15) (0.09, 0.12, 0.16, 0.20) (0.12, 0.16, 0.2, 0.25) (0.09, 0.09, 0.135, 0.135) (0.045, 0.075, 0.1, 0.14) (0.06, 0.1, 0.125, 0.175) Step 3. Determine the fuzzy positive-ideal solution X + and fuzzy negative-ideal solution X, respectively, X + = [(0.15, 0.15, 0.20, 0.20), (0.1, 0.1, 0.15, 0.15), (0.1, 0.1, 0.15, 0.15), (0.1, 0.1, 0.15, 0.15), (0.09, 0.12, 0.16, 0.20), (0.14, 0.18, 0.225, 0.25)] X = [(0.108, 0.108, 0.144, 0.144), (0.056, 0.056, 0.084, 0.084), (0.086, 0.086, 0.129, 0.129), (0.069, 0.069, , ), (0, 0.03, 0.04, 0.08), (0.06, 0.1, 0.125, 0.175)] Step 4. Calculate the distance of each alternative from fuzzy positive-ideal solution and fuzzy negative-ideal solution, respectively, as Table 2.

30 24 Qiuping Wang and Daohong Zhang No. 3 Table 2: The distance measurement X 1 X 2 X 3 X 4 D D as Step 5. Calculate the relative closeness coefficient of each alternative to the ideal solution C 1 = , C 2 = , C 3 = , C 4 = Step 6. According to the closeness coefficient in the descending order, the ranking order of four alternatives can be determined as follows: X 3 X 1 X 4 X 2. Therefore, the best selection is X 3. The example result by method of this paper proposal is consistent with result by method of fuzzy compromise decision method which is first synthesis, then weighted decision making pattern introduced in reference [10]. Generally speaking, except some decision-making situations with quite obvious differences, the result which the different method obtains isn t always consistent. Therefore in the actual decision-making, we should use several different methods to solve, synthesize and compare result from each method in order to obtain a relatively reliable conclusion. 5. Conclusion This paper has proven the product of interval number and the triangular fuzzy number is a trapezoidal fuzzy number. We have proposed new method for multiple attribute decision making problem with interval number attribute weight and triangular fuzzy number attribute value. A numerical example is used to illustrate the method. Through the example, we can find this method is reliable, the thinking is clear and is one practical decision making method. References [1] Fengji Gao, Yongfang Huang, Xiaoyan Ren, Multiple Attribute Interval Decision Making Method by Similarity to Ideal Point, The Journal of Quantitative & Technical Economics (in Chinese), 18(2001), No.2, [2] Jijun Zhang, Yuying Fan, TOPSIS of Multple Attribute Decision Making Promlem With Interval Valued Weight, Systems Engineering and Electronics (in Chinese), 24(2002), No.11, [3] Zeshui Xu, A Method for Multi-objective Decision Making with Incomplete Information, Operations Research and Management Science (in Chinese), 10(2001), No.2, [4] Jijun Zhang, Jiacai Liu, The Study of Decision Making Method to Interval Number Multiple Attribute Decision Making Problem, FORECASTING (in Chinese), 21(2002), No.1,

31 Vol. 4 A new decision making method for fuzzy multiple attribute decision making problem 25 [5] Qingli Da, Zeshui Xu, Single-objective optimization model in uncertain multi-attribute decision-making, Journal of Systems Engineering (in Chinese), 17(2002), No.1, [6] Quan Zhang, Zhi ping Fan, Dehui Pan, A Ranking Approach for Interval Numbers in Uncertain Multiple Attribute Decision Making Problems, Systems Engineering-Theory & Practice (in Chinese), 19(1999), No.5, [7] Zhiping Fan, Quan Zhang, The Revision for the Uncertain Multiple Attribute Decision Making Models, Systems Engineering-Theory & Practice, 19(1999), No.12, [8] Vania Peneva, Ivan Popchev, Comparison of clusters from fuzzy numbers, Fuzzy Sets and Systems, 97(1998), No.1, [9] Chen-Tung Chen, Extensions of the TOPSIS for group decision-making under fuzzy environment, Fuzzy sets and systems, 114(2000), 1-9. [10] Rongjun Li, Fuzzy Multiple-criteria Decision-making Theory and Application, Science and Technology Press, Beijing (in Chinese), 2002.

32 Scientia Magna Vol. 4 (2008), No. 3, A new Smarandache function and its elementary properties Jing Fu and Yu Wang Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any positive integer n, we define a new Smarandache function G(n) as the m smallest positive integer m such that φ(k) is divisible by n, where φ(n) is the Euler function. k=1 The main purpose of this paper is using the elementary methods to study the elementary properties of G(n), and give three interesting formulas for it. Keywords A new Smarandache function, series, inequality. 1. Introduction and results For any positive integer n, we define a new Smarandache function G(n) as the smallest m positive integer m such that φ(k) is divisible by n. That is, k=1 G(n) = min{m : n m φ(k), m N}, k=1 where φ(n) is the Euler function, N denotes the set of all positive integers. For example, the first few values of G(n) are: G(3) = 7, G(4) = 4, G(5) = 11, G(6) = 7, G(7) = 29, G(8) = 5, G(9) = 9, G(10) = 11, G(11) = 23, G(12) = 7, G(13) = 43, G(14) = 29, G(15) = 11, G(16) = 5. About the properties of this function, it seems that none had studied it yet, at last we have not seen any related papers before. Recently, Professor Zhang Wenpeng asked us to study the arithmetical properties of G(n). The main purpose of this paper is using the elementary methods to study this problem, and prove the following three conclusions: Theorem 1. For any prime p, we have the calculating formulae G(p) = min{p 2, q(p, 1)}; G(p 2 ) = q(p, 2), if q(p, 2) < p 2 ; G(p 2 ) = p 2, if q(p, 1) < p 2 < q(p, 2); G(p 2 ) = q(p, 1), if p 2 < q(p, 1) < 2p 2 ; and G(p 2 ) = 2p 2, if q(p, 1) > 2p 2, where q(p, i) is the i-th prime in the arithmetical series {np + 1}.

33 Vol. 4 A new Smarandache function and its elementary properties 27 Theorem 2. G(n) is a Smarandache multiplicative function, and moreover, the Dirichlet G(n) series n 2 is divergent. n=1 Theorem 3. Let k 2 be a fixed positive integer, then for any positive integer group (m 1, m 2,, m k ), we have the inequality G(m 1 m 2 m k ) G(m 1 )G(m 2 ) G(m k ). Note. For any positive integer n, we found that n = 1, 4, 9 are three positive integer solutions of the equation G(n) = n. Whether there exist infinite positive integers n such that the equation G(n) = n is an interesting problem. We conjecture that the equation G(n) = n has only three positive integer solutions n = 1, 4, 9. This is an unsolved problem. 2. Proof of the theorem In this section, we shall complete the proof of our theorems directely. First we prove Theorem 1. Let G(p) = m, namely p φ(k), p φ(k), 0 < s < m. If m = q α1 1 qα2 2 qαr r be m s the k=1 factorization of m into prime powers, then φ(m) = q α1 1 1 (q 1 1)q α2 1 2 (q 2 1) q αr 1 from the definition of G(n) we can deduce that p divides one of φ(q αi i q αi 1 i k=1 r (q r 1), ), 1 i r. So p divides or q i 1. We discuss it in the following two cases: (i). If p q i 1, then we must have m = q i = lp+1 be the smallest prime in the arithmetical series {kp + 1}. (ii). If p q αi 1 i, then we have have α i = 2, m = qi 2 = p2. Combining (i) and (ii) we may immediately deduce that G(p) = min{p 2, lp + 1}, where lp + 1 be the smallest prime in the arithmetical series {kp + 1}. Similarly, we can also deduce the calculating formulae for G(p 2 ). This proves Theorem 1. Now we prove Theorem 2. For any positive integer n > 1, let n = p α1 1 pα2 2 pαr r be the factorization of n into prime powers, G(p αi then from the definition of G(n) we have p αi i i ) = m i, 1 i r, m = max{m 1, m 2,, m r }, divides φ(1)φ(2) φ(m i ) for all 1 i r. So p αi i divides divides φ(1)φ(2) φ(m). Since (p i, p j ) = 1, i j, so we must have n = p α1 1 pα2 2 pαr r φ(1)φ(2) φ(m). Therefore, G(n) = m = max{g(p α1 1 ), G(pα2 2 ),, G(pαr r Smarandache multiplicative function. From Theorem 1 we may immediately get n=1 G(n) n 2 > p G(p) p 2 > p 1 p =. )}. So G(n) is a G(n) So the Dirichlet series n 2 is divergent. n=1 Finally, we prove Theorem 3. First we prove that the inequality G(m 1 m 2 ) G(m 1 )G(m 2 ) holds for any positive integer m 1 and m 2. Let G(m 1 ) = u, G(m 2 ) = v, from the definition of G(n) we can easily get u v m 1 φ(i), m 2 φ(i). i=1 i=1

34 28 Jing Fu and Yu Wang No. 3 Without loss of generality we can assume u v, then uv k=1 Notice that φ(k) = u φ(k) k=1 uv k=u+1 φ(k) = u φ(k) φ(u + 1) φ(v) φ(2v) φ(uv) k=1 φ(2) φ(2v), φ(3) φ(3v),, φ(u) φ(uv), φ(u + 1) φ(u + 1),, φ(v) φ(v), this means that or Hence i=1 v φ(i) i=1 uv k=u+1 u v φ(i) φ(i) i=1 m 1 m 2 uv k=1 φ(k), uv k=1 φ(k). φ(k). From the definition of G(n) we know that G(m 1 m 2 ) uv, or G(m 1 m 2 ) G(m 1 )G(m 2 ). If k 3, then applying the above conclusion we can deduce that G(m 1 m 2 m k ) = G(m 1 (m 2 m k )) G(m 1 )G(m 2 m k ) G(m 1 )G(m 2 )G(m 3 m k ) G(m 1 )G(m 2 ) G(m k ). This completes the proof of Theorem 3. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [3] Pan Chengdong and Pan Chengbiao, The elementary number theory, Beijing University Press, Beijing, [4] Zhang Wenpeng, The elementary number theory, Shaanxi Normal University Press, Xi an, [5] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, 1996.

35 Scientia Magna Vol. 4 (2008), No. 3, On some congruent properties of the Smarandache numerical carpet Yanlin Zhao Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract In this paper, we use the elementary methods to study some divisibility of the Smarandache numerical carpet, and obtain two interesting congruent theorems. Keywords Divisibility, Smarandache numerical carpet, elementary method. 1. Introduction and results In his book Only problems, not solutions, F.Smarandache introduced many interesting sequences and arithmetical functions, at the same time, he also proposed 105 unsolved problems and conjectures. One of the problems is that he asked us to study the properties of the numerical carpet. It has the general form 1 1 a 1 1 a b a 1 1 a b c b a 1 1 a b c d c b a 1 1 a b c d e d c b a 1 1 a b c d c b a 1 1 a b c b a 1 1 a b a 1 1 a 1 1 On the border of level 0, the elements are equal to 1, they form rhombus. Next, on the border of level 1, the elements are equal to a, where a is the sum of all elements of the previous border, the a s form a rhombus too inside the previous one. Next again, on the border of level 2, the elements are equal to b, where b is the sum of all elements of the previous border, the b s form a rhombus too inside the previous one. And so on

36 30 Yanlin Zhao No. 3 It can be described as a 1... a a 1 a 2... a 2 a a 1 a 2 a 3... a 3 a 2 a a 1 a 2... a 2 a a 1... a Let n, k be integers with n 0 and k 0, the n th square can be presented as a set of sub-squares. In [2], Mladen and T. Krassimir gave and proved the general formula: where integer k 2. k 2 a k = 4(k + 1) (4n + 1 4i), i=0 Let C(n, k) denotes a k in n-th square, then C(n, k) = 1, for k = 0; C(n, k) = 4(n + 1), for k = 1; and C(n, k) = 4(n + 1) k 2 (4n + 1 4i), for 2 k n + 1. i=0 The main purpose of this paper is using the elementary methods to study the divisibility properties of the sequence {C(n, k)}, and give two interesting congruent theorems. That is, we shall prove the following conclusions: Theorem 1. Let n be a positive integer. Then for any prime p with 4n 4k+9 p 4n+1 and p 1 mod 4, we have the congruence C(n, k) 0 mod p. [ ] n k + 2 Theorem 2. Let n be a positive integer, p be any prime with n k [ ] 3 n + 1 p n + and p 3 mod 4, then we have the congruence 3 C(n, k) 0 mod p.

37 Vol. 4 Some divisibility of Smarandache numerical carpet Proof of the theorems In this section, we shall use the elementary methods to complete the proof of the theorems. First we prove Theorem 1. For any positive integer n and prime p 1 mod 4, if 4n 4k + 9 p 4n + 1, then it is clear that 4n 4k + 9 4n + 1 or k 2, from this we have k 2 C(n, k) = 4(n + 1) (4n 4i + 1) i=0 k 2 = 4(n + 1) (4(n i) + 1). It is obviously that C(n, 2) = 4(n + 1)(4n + 1), so if k = 2, then from 4n 4k + 9 p 4n + 1 we know that p = 4n + 1, and C(n, 2) 0 mod p. This means that Theorem 1 is true. Now we assume that 2 < k n, so that k 2 C(n, k) = 4(n + 1) i=0 (n + 1) 4 k i=0 (4(n i) + 1) (n + 1) 4 k k 2 i=0 ( n i + 3p + 1 ) 4 k 2 i=0 ( n i + 4 ) mod p mod p. (1) where x denotes the solution of the congruence equation x x 1 mod p, and 4 = 3p Since 4n 4k + 9 p 4n + 1, so there must exist an integer 0 i k 2 such that p = n i + 3p + 1. Therefore, from (1) we may immediately deduce that 4 k 2 C(n, k) = 4(n + 1) (4(n i) + 1) 0 mod p. This proves Theorem 1. Now we prove Theorem 2. Let n be a positive integer, p be a prime with [ ] [ ] n k + 2 n + 1 n k p n + and p 3 mod Note that 4 = p + 1, then from the definition of C(n, k) we have 4 i=0 k 2 C(n, k) = 4(n + 1) (4(n i) + 1) (n + 1)4 i=0 ( p + 1 (n + 1) 4 ) k 2 k 2 i=0 From congruence (2) we know that if n k+2+ p The inequality n k p p n + p n k n k 3 k 2 k 2 ( n i + p ( ) n i + 4 mod p i=0 ) mod p. (2) p n+ p + 1, then C(n, k) 0 mod p. 4 implies p n + n

38 32 Yanlin Zhao No. 3 Since prime p must be a positive integer, so we have [ ] [ ] n k + 2 n + 1 n k p n This completes the proof of Theorem 2. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan publishing House, [2] V. Mladen and T. Krassimir, Remarks on some of the Smarandache s problem. part 2, Scientia Magna, 1(2005), No.2, [3] F. Smarandache, Sequences of numbers involved in unsolved problems, Hexis, [4] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, [5] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [6] Liu Yanni, Li Ling and Liu Baoli, Smarandache unsolved problems and new progress (in Chinese), High American Press, [7] Yi Yuan and Kang Xiaoyu, Research on Smarandache problems (in Chinese), High American Press, [8] Chen Guohui, New progress on Smarandache problems (in Chinese), High American Press, 2007.

39 Scientia Magna Vol. 4 (2008), No. 3, On the Smarandache prime part sequences and its two conjectures 1 Yahui Yu and Lixiang Cai Luoyang Institute of Science and Technology, Luoyang, Henan, P.R.China Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any positive integer n 2, the Smarandache Inferior Prime Part {p p(n)} is defined as the largest prime less than or equal to n; The Smarandache Superior Prime Part {P p(n)} is defined as the smallest prime greater than or equal to n. In this paper, we defined some determinants involving the Smarandache prime part sequences, and introduced two conjectures proposed by professor Zhang Wenpeng. Keywords Smarandache prime part sequences, Determinant, Conjecture. 1. Introduction For any positive integer n 2, the Smarandache Inferior prime part {p p (n)} is defined as the largest prime less than or equal to n. The Smarandache Superior prime part {P p (n)} as the smallest prime greater than or equal to n. For example, the first few values of these sequences are p p (2) = 2, p p (3) = 3, p p (4) = 3, p p (5) = 5, p p (6) = 5, p p (7) = 7, p p (8) = 7, p p (9) = 7, p p (10) = 7, p p (11) = 11, p p (12) = 11, p p (13) = 13, p p (14) = 13, p p (15) = 13, p p (16) = 13, p p (17) = 17, p p (18) = 17, p p (19) = 19,. P p (1) = 2, P p (2) = 2, P p (3) = 3, P p (4) = 5, P p (5) = 5, P p (6) = 7, P p (7) = 7, P p (8) = 11, P p (9) = 11, P p (10) = 11, P p (11) = 11, P p (12) = 13, P p (13) = 13, P p (14) = 17, P p (15) = 17, P p (16) = 17, P p (17) = 17, P p (18) = 19, P p (19) = 19,. From the definitions of these sequences we know that for any prime q, we have p p (q) = P p (q) = q. In his books Only problem, Not solutions [1] and Sequences of Numbers Involved in Unsolved Problem [2], Professor F.Smarandache asked us to study the properties of these sequences. About these two problems, some authors had studied them, and obtained some interesting results, see references [3], [4], [5] and [6]. For example, Xiaoxia Yan [6] studied the the asymptotic properties of S n I n, and proved that for any positive integer n > 1, we have the asymptotic formula S n I n = 1 + O(n 1 3 ), where I n = {p p (2) + p p (3) + + p p (n)}/n and S n = {P p (2) + P p (3) + + P p (n)}/n. 1 This work is supported by the Shaanxi Provincial Education Department Foundation 08JK433.

40 34 Yahui Yu and Lixiang Cai No. 3 The sequences {p p (n)} and {P p (n)} are very interesting and important, because there are some close relationships between the Smarandache prime part sequences and the prime distribution problem. Now we introduced two determinants formed by the Smarandache prime part sequences. They are defined as follows: For any positive integer n, c(n), C(n) are n n determinants, namely p p (2) p p (3) p p (n + 1) p p (n + 2) p p (n + 3) p p (2n + 1) c(n) = p p (n(n 1) + 2) p p (n(n 1) + 3) p p (n 2 + 1) and C(n) = P p (1) P p (2) P p (n) P p (n + 1) P p (n + 3) P p (2n) P p (n(n 1) + 1) P p (n(n 1) + 2) P p (n 2 ) For example, by definitions and calculating, we can find some values of these determinants as the following table: n c(n) C(n) n c(n) C(n) About the elementary properties of these determinants, it seems that none had studied them, at least we haven t seen any related papers before. Recently, Professor Zhang Wenpeng

41 Vol. 4 On the Smarandache prime part sequences and its two conjectures 35 asked us to study the properties of these determinants, at the same time, he also proposed following two conjectures: Conjecture 1. For any composite number n 6, we have the identities c(n) = 0 and C(n) = 0; Conjecture 2. For any prime q, we have c(q) 0 and C(q) 0. From the above table, we believe that these two conjectures are correct. About Conjecture 1, we have solved it completely, which will be published in Pure and Applied Mathematics. But for Conjecture 2, it s still an open problem. We think that Conjecture 2 is very interesting and important, since if it is true, then we can get a new discriminant method to distinguish prime from integer numbers through calculating these kinds of determinants. So we introduce the conjectures in this place, and hope more scholars who are interested in to study it with us. References [1] F. Smarandache, Only Problem, Not Solutions, Chicago, Xiquan Publishing House, [2] F. Smarandache, Sequences of Numbers Involved in Unsolved Problems, Hexis, [3] D. R. Heath-Brown, The differences between consecutive primes, Journal of London Math. Soc., 18(1978), No.2, [4] D. R. Heath-Brown, The differences between consecutive primes, Journal of London Math. Soc., 19(1979), No.2, [5] Wu Qibin, A composite function involving the Smarandache function, Pure and Applied Mathematics, 23(2007), No.4, [6] Xiaoxia Yan, On the Smarandache prime part, Scientia Magna, 3(2007), No.3, [7] Yi Yuan and Kang Xiaoyu, Research on Smarandache Problems, High American Press, [8] Chen Guohui, New Progress On Smarandache Problems, High American Press, [9] Liu Yanni, Li Ling and Liu Baoli, Smarandache Unsolved Problems and New Progress, High American Press, 2008.

42 Scientia Magna Vol. 4(2008), No. 3, On the continued fractions and Dirichlet L-functions 1 Shanghai Jin and Ping Tang Department of Applied Mathematics, Xi an University of Technology, Xi an, Shaanxi, P.R.China Abstract The main purpose of this paper is using the elementary methods to study the relationship between the continued fractions and Dirichlet L-functions, and give some interesting identities. Keywords Continued fractions, Dirichlet L-functions, identities. 1. Introduction Let x 0, x 1, x 2, be integers with x j 1, j 1. We can express the original fraction p n 1 in the form x 0 + 1, n 0. Such an expression is called a continued fraction, and x 1+ x xn 1 we denote it by x 0, x 1,, x n. About the characteristic properties of continued fractions, it can be found in the references [1] and [2]. In this paper, we shall study the relationship between the Dirichlet L-functions and the continued fractions, and obtain some interesting identities. That is, we shall prove the following main conclusion: Theorem. Let p be a prime, a be a positive integer with (a, p) = 1 and 1 a < p. Then we have the identity χ mod p,χ( 1)= 1 = π2 (p 1) 12p χ(a) L(1, χ) 2 [ n ] ( 1) k+1 x k + ( 1) n 1 q n 1 + a q n p ( 1 + ( 1)n ), k=0 where χ be the Dirichlet character mod p, L(1, χ) be the Dirichlet L-function corresponding character χ and a/p = p n /q n = x 0, x 1,, x n. For a = 2, 3,4 and p 1, from the above Theorem and the definition of continued fractions 2 we may immediately obtain the following four Corollaries. 1 This work is supported by the Shaanxi Provincial Education Department Foundation (08JK398) of P.R.China q n

43 Vol. 4 On the continued fractions and Dirichlet L-functions 37 Corollary 1. Let k > 1 be an odd prime, then we have the identity χ(2) L(1, χ) 2 = π2 φ 2 [ (k) 1 12 k 2 3 ]. k χ mod k,χ( 1)= 1 Corollary 2. Let p be an odd prime and χ be the Dirichlet character mod p. Then π 2 χ(3) L(1, χ) 2 = 36 (p 1)2 (p 10) p 2, if p 1 mod 3; π χ mod p,χ( 1)= 1 2 (p 1)(p 2)(p 5) 36 p 2, if p 2 mod 3. Corollary 3. Let p be an odd prime and χ be the Dirichlet character mod p. Then π 2 χ(4) L(1, χ) 2 = 48 (p 1)2 (p 17) p 2, if p 1 mod 4; π χ mod p,χ( 1)= (p 1)(p2 6p + 17) p 2, if p 3 mod 4. Corollary 4. Let p be an odd prime and χ be the Dirichlet character mod p. Then ( ) p 1 χ L(1, χ) 2 = π2 (p 1)(p 2 4p + 5) 2 24p 2. χ mod p,χ( 1)= 1 It is clear that these four Corollaries are an extension of Walum [3]. 2. Some Lemmas To complete the proof of the Theorem, we need the following three Lemmas. Lemma 1. Let x 0, x 1,, x n be integers and p i, q i (0 i n) be natural numbers, continued fraction x 0, x 1,, x i = p i q i. Then p i q i 1 p i 1 q i = ( 1) i+1, 1 i n. and p 2 = 0, p 1 = 1, q 2 = 1, q 1 = 0, p i = x i p i 1 + p i 2, q i = x i q i 1 + q i 2, 0 i n. Proof. (See reference[1] ). Lemma 2. Let k and h be integers with k 3 and (h, k) = 1. Then we have S(h, k) = 1 d 2 π 2 k φ(d) d k χ mod d,χ( 1)= 1 χ(h) L(1, χ) 2, q (( )) (( )) a ah where S(h, q) = is the Dedekind sum and φ(k) is the Euler s function. q q a=1 Proof. (See reference [4]). Lemma 3. Let a and q be two positive integers with (a, q) = 1. Then we have the identity [ S(a, q) = 1 n ] ( 1) k+1 x k + ( 1) n 1 q n 1 + a + 12 q n q k=0 1 + ( 1)n, 8

44 38 Shanghai Jin and Ping Tang No. 3 where a/q = p n /q n = x 0, x 1,, x n. Proof. Let continued fraction x 0, x 1,, x n = p n q n successive terms p n q n and p n 1 q n 1, from Lemma 1 we know that and suppose p n = a. For the q n q p n q n 1 q n p n 1 = ( 1) n+1, n 1. (1) Using the properties of Dedekind sums and (1) we get S(p n, q n ) = S(p n q n 1 q n 1, q n ) (2) = S ( q n 1 (( 1) n+1 ) + p n 1 q n ), q n = S ( ( 1) n+1 ) q n 1, q n = ( 1) n+1 S (q n 1, q n ) = ( 1) n+1 S(q n 1, q n ), where q n 1 denotes the solution x of the congruence equation xq n 1 1( mod q n ). From the reciprocity formula and Lemma 1 we obtain S(q n 1, q n ) + S(q n, q n 1 ) = q2 n + qn q n q n 1 4 = 1 ( qn 1 + q ) n q n q n 1 12q n q n 1 4 = 1 [( qn 1 + q ) n 2 + x n 12 q n 1 Hence, by expression (3) and Lemma 1 we obtain q n + ( 1) n+1 ( pn q n p n 1 q n 1 )] 1 4. S(q n 1, q n ) (4) = S(q n, q n 1 ) + 1 [( qn 1 + q ) ( n 2 + x n + ( 1) n+1 pn p )] n q n q n 1 q n q n 1 4 = S(q n 2, q n 1 ) + 1 [( qn 1 + q ) ( n 2 + x n + ( 1) n+1 pn p )] n q n q n 1 q n q n 1 4 [( ) ( qn 2 + ( 1) n pn 1 )] = S(q n 3, q n 2 ) 1 + q n 3 + x n 1 p n 2 12 q n 1 q n 2 q n 1 q n [( qn 1 + q ) ( n 2 + x n + ( 1) n+1 pn p )] n q n q n 1 q n q n 1 4 = S(q n 3, q n 2 ) + 1 [( qn 1 q ) ( n 3 + x n x n 1 + ( 1) n+1 pn p n 2 12 q n q n 2 q n q n 2 If n is an even number, then from (4) we have S(q n 1, q n ) = S(q 1, q 0 ) = 1 12 i=0 [ n = 1 12 [ n 1 i=0 [ (qn 1 q n ( 1) i x n i + q n 1 q n ( 1) i x n i + q n 1 q n q ) ( 1 + ( 1) n+1 pn q 0 q n p 0 + ( 1) n+1 p n q n + ( 1) n p 0 + ( 1) n+1 p n q n ]. q 0 q 0 ] )]. (3) ) ] n 1 + ( 1) i x n i (5) i=0

45 Vol. 4 On the continued fractions and Dirichlet L-functions 39 If n is an odd number, then from (4) we have [ (qn 1 S(q n 1, q n ) = S(q 0, q 1 ) + 1 q 0 12 q n q 1 [ = S(1, x 1 ) + 1 (qn q n x 1 ) + ( 1) n+1 ( pn p 1 q n q 1 ) ( + ( 1) n+1 pn x 0x q n x 1 [ q n 1 + ( 1) n+1 p n q n ] = (x 1 1)(x 1 2) x 1 12 [ = 1 n ( 1) i x n i + q n 1 + ( 1) n+1 p n 12 q i=0 n q n Now combining (2), (5) and (6) we obtain [ S(p n, q n ) = 1 n ( 1) k+1 x k + ( 1) n 1 q n 1 12 k=0 q n ) ] n 2 + ( 1) i x n i i=0 2 n 2 x 0 + q n x p n q n Note that p n /q n = a/q we may immediately get [ S(a, q) = 1 n ] ( 1) k+1 x k + ( 1) n 1 q n 1 + a + 12 q k=0 This completes the proof of Lemma 3. q n ] + ) n 2 + i=0 i=0 1 + ( 1)n ( 1)n. 8 (6) ( 1) i x n i ] ( 1) i x n i ] 3. Proof of the theorem In this section, we shall complete the proof of the theorem. Using Lemma 2 we have χ mod p,χ( 1)= 1 Then from Lemma 3 and expression (7) we can easily get χ(a) L(1, χ) 2 χ mod p,χ( 1)= 1 = π2 (p 1) p = π2 (p 1) 12p χ(a) L(1, χ) 2 = π2 (p 1) S(a, p). (7) p [ ( n ) ] 1 ( 1) k+1 x k + ( 1) n 1 q n 1 + a 1 + ( 1)n + 12 q n p 8 k=0 [ n ] ( 1) k+1 x k + ( 1) n 1 q n 1 + a q n p ( 1 + ( 1)n ). k=0 This completes the proof of Theorem. The proof of the Corollaries. For any integer k > 1 and (h, k) = 1, using the Möbius inversion formula and Lemma 2 we have χ mod k,χ( 1)= 1 χ(h) L(1, χ) 2 = π 2 φ(k) k d k µ(d) d S(h, k ). (8) d

46 40 Shanghai Jin and Ping Tang No. 3 Taking h = 2 in (8), note that S(2, 1) = 0 and the continued fraction 2 k/d = p 2 = 0, k/d 1, 2 = x 0, x 1, x 2. q 2 2 If k be a odd prime, then from (8) and Theorem we have χ mod k,χ( 1)= 1 = π 2 φ(k) k = π2 φ(k) 12k = π2 φ(k) 12k = π2 φ 2 (k) 12 k d k,d<k d k,d<k d k,d<k [ k χ(2) L(1, χ) 2 = π 2 φ(k) k d k µ(d) d S(2, k d ) µ(d) d S(2, k d ) [ 2 ] µ(d) ( 1) k+1 x k + ( 1) q d q 2 p k=0 ( µ(d) k/d 1 2 k/d ) d 2 k/d k/d ]. This proves Corollary 1. Using Theorem and the definition of continued fraction we can also deduce that and χ χ mod p,χ( 1)= 1 mod p,χ( 1)= 1 χ χ(3) L(1, χ) 2 = χ(4) L(1, χ) 2 = mod p,χ( 1)= 1 This completes the proof of Corollary 2, 3 and 4. π 2 36 (p 1)2 (p 10) p 2, if p 1 mod 3; π 2 (p 1)(p 2)(p 5) 36 p 2, if p 2 mod 3. π 2 48 (p 1)2 (p 17) p 2, if p 1 mod 4; π 2 48 (p 1)(p2 6p + 17) p 2, if p 3 mod 4. ( ) p 1 χ L(1, χ) 2 = π2 (p 1)(p 2 4p + 5) 2 24p 2. References [1] Pan Chengdong and Pan Chengbiao, Elementary Number Theory, Beijing University, Beijing, 1992, [2] G. H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers, Clarendon, Oxford, [3] H. Walum, An exact formula for an average of L-series Illinois J. Math, 26(1982), 1-3. [4] Wenpeng Zhang, On the mean values of Dedekind sums, Journal De Theorie des Nombres de Bordeaux, 8(1996),

47 Scientia Magna Vol. 4 (2008), No. 3, Research on the connectivity of public transportation network based on set method Chunming Liu, Wansheng He, Hongming Xia and Zhongshe Gao Department of Computer Science and Technology, Tianshui Normal University Tianshui, Gansu, , P. R. China School of Mathematics and Statistics,Tianshui Normal University Tianshui, Gansu, , P. R. China cucmeLiu@gmail.com Abstract A connectivity judgment method for public transportation network based on set idea is studied in this article. Set is successfully applied to solve the problem of connectivity judgment on multi-vertex and multi-edge situation. Search level control strategy is adapted to control the search level of route, and three buffer tables are designed to improve the search speed and reduce the complexity of the algorithm. Keywords Traffic management, connectivity, transportation network, search level control. 1. Introduction Public transport is the main travel way for urban civilians and it is also a way advocated by many governments and organizations under the environmental protection concept. On one hand, along with the rapid construction of urban traffic, public travel becomes more and more convenience; on the other hand, problems come forth with multi-route selection. The civilians had to balance among fee, time cost, distance and some other areas, especially those in large cities or metropolis. And all this must be built on the basis of the pathway connectivity judgment, which are the major problem, as well as the common pathway connectivity judgment and optimization strategy under the circumstances of huge number of bus stations and bus lines, discussed in this article. 2. Some known results 2.1. Problem description The pathway connectivity discussed in this article means the connectivity between one station and another, directly or through a third or more other stations, then find out the routes and the shortest one if they are interlinked. Basically, there are three types of public transport network distribution which include, all stations on the up-run line and down-run line are the same (also called Bidirectional Lines),

48 42 Chunming Liu, Wansheng He, Hongming Xia and Zhongshe Gao No. 3 part of the stations are the same (also called Unilines) and Loop Lines. Meanwhile there are two types of fee system known as Flat Fare and Sectional Fare. Following are three lines information of a city s public transport system. l 135, Bidirectional Line; Sectional Fare. s 0417 s 0272 s 3425 s 3476 s 2974 s 0234 s 0521 s 3806 s 1682 s 3559 s 0928 s 3564 s 0079 s 3175 s 2866 s 3172 s 3269 s 3603 s 1633 s 2578 s 2579 s 2576 s 2577 l 181, Uniline; Flat Fare. Up-Run Line: s 0238 s 0540 s 0542 s 3024 s 0494 s 0973 s 3077 s 2082 s 2213 s 2210 s 3332 s 3351 s 1414 s 0228 s 0916 s 2811 s 0019 s 0966 Down-Run Line: s 0966 s 0019 s 2811 s 0916 s 0228 s 1414 s 3351 s 3332 s 2210 s 2213 s 2082 s 3077 s 0973 s 0494 s 3024 s 0542 s 0541 s 0152 s 1765 s 3506 s 0238 l 590, Loop Line; Flat Fare. s 3295 s 3265 s 1334 s 1333 s 2285 s 1341 s 0327 s 0317 s 0325 s 0532 s 1734 s 0314 s 1749 s 1831 s 1337 s 0705 s 0704 s 3295 Where l n is the n-th line s No. of the city and s m is the m-th station of the city Specialties of the problem Algorithm for connectivity of graph (include Breadth First Search and Depth First Search), adjacency matrix and genetic algorithm are all popular methods adopted to judge pathway connectivity [1]. Outwardly, public transport network can be convert to a digraph whose vertexes are stations and edges are lines between two stations. However there are actually more than one line that go through each station in the transport network, namely, there are unfixed number of edges between two adjoining vertexes. Therefore, the commonly used router algorithms for connectivity of graph are inapplicable since the time complexity and space complexity increase rapidly, for instance, there are more than 850 bus lines and more than 6400 bus stations in city Beijing [2], since the transport network is a n-th Iterated Line Digraphs [3]. Moreover, on one hand, because of the characteristics of intersecting and overlapping, there are always more than one line between two adjoining stations. On the other hand, even if there are route between two stations, it may need ongoing transferring, however, it is contrary to the public s travel habits which are the less transferring, the better. According to the experience of livelihood, the public trend to reduce their transferring frequency when other factors (such as cost, time, etc.) difference is not very significant. That is, if there are more than one route between two stations, direct line has priority over transferring and transferring once has priority over transferring more times and so on, and other factors will be taken into considering only when transferring frequencies are equal. Therefore, the following discussing was based on the public s travel habits. 3. Set method for connectivity 3.1. Set description

49 Vol. 4 Research on the connectivity of public transportation network based on set method 43 Provide elements l i, s j and sets L i, S j, A j, B (j,i), which are l i : Element that indicate line No. i; s j : Element that indicate station No. j; L i : Set of all lines that pass through station s i ; S j : Set of stations that line l j pass through; A j : Set of stations that can be reached start from s j ; B (j,i) : Set of all stations that can be reached with line l i start from station s j ; The following expressions can be drown out: 1). Direct Line: the following expression is obtained if there have direct line run from start station s start to destination station s dest : s start s dest {s start, s dest } A start A dest. (1) Where s start s dest indicate there are direct line run from s start to s dest. Therefore, when {s start, s dest } A start A dest is establish, namely, there are direct line run from s start to s dest, the collection C of all direct lines are the intersection of set L start whose elements are lines that run through s start and set L dest whose elements are lines that run through s dest : C = L start L dest. (2) 2). Transfer Once: The circumstance of transfer once from s start to s dest can be converted to the iterate of two direct lines, that is, the direct line from the start station s start to the transfer station s transfer and the direct line from the transfer station s transfer to the destination station s dest. Therefore the set D of transfer stations is: D = A start A dest. (3) 3). Transfer multi-times: Similarly, The circumstance of transfer n-times from s start to s dest can be converted to the iterate of (n + 1) direct lines. Direct line and transfer-needed circumstances are schematically shown in Fig.1. Figure 1: The schematic view of (a) direct line, (b) transfer once and (c) transfer twice of two stations 3.2. Optimization of algorithm

50 44 Chunming Liu, Wansheng He, Hongming Xia and Zhongshe Gao No. 3 As has been analyzed that the public always trend to choose routes with less transferring frequencies, therefore, direct lines should be searched in top priority; deepen search level when direct lines are not available, and so on and so forth, until searching accomplished (succeed or failed) [4]. In addition, three memory tables known as line-station hash table, line-station searching linked-list and station-line searching linked-list are designed to improve the searching speed and reduce the searching complexity Route searching strategy Search-level control strategy was introduce to integrate the Breadth-First Search method (BFS) and Depth-First Search method (DFS). The concrete steps are list in the following: 1). Construct set L start, L dest, A start and A dest for s start and s dest. 2). Figure out the intersection of L start and L dest, L start L dest. Then find out all identical elements in both set L start and set L dest via BFS, thus the result C is the set of all direct lines. a). If C is not empty, record all elements in C which are all direct lines. b). If C is empty, deepen the search levels and continue with step 3); 3). Figure out the intersection of A start and A dest, A start A dest. Then find out all identical elements in both set A start and set A dest via DFS, thus the result set D is the set of all transfer stations. a). If D is not empty, record all elements in D which are all transfer stations. Then search direct lines with s start and D, s dest and D, respectively. b). If D is empty, keep on deepening the search levels Hash table structure of line-station Hash table structure of line-station stores the relationship of lines and stations. When the hash value, which are corresponded a line and a station, is 1, the corresponding line runs through the corresponding station, and 0, doesn t. The hash table structure of line-station is schematically shown in Fig. 2. s 1 s 2 s 3 s n 1 s n l l l l m l m Figure 2: Schematic view of the hash table structure of line-station As can be clearly seen from Fig. 2 that row direction of the hash table represents the lines and column direction, the stations. Moreover, the hash value 1 in row direction marks all stations that the corresponding line runs through, and in column direction marks all lines that run through the corresponding station. As is known to all that the time complexity of search algorithm using hash table structure is the constant of O(1) [5].

51 Vol. 4 Research on the connectivity of public transportation network based on set method 45 as: Two important logic relationships are implicated in the above hast table structure, known 1). There are direct line between every two stations whose hash value is 1 in the same row. 2). Transfer is available between every two lines whose hash value is 1 in the same column. Actually, the purpose of constructing the hash table of line-station is to perform high speed search utilizing these two relationships Line-station searching linked-list Line-station searching linked-list stores the stations that every line runs through successively in a duplex linked-list. The list is sorted by line NO. to locates elements with binary search. Line-station search list structure is schematically shown in Fig. 3. Figure 3: Schematic view of line-station searching linked-list structure Since the time complexity of binary search is O(log(n)), the total time complexity of the line-station duplex searching linked-list is O(log(m)) + O(log(max(spl))), (4) where m is the count of all lines; max(spl) is the max count of stations that a line runs through Station-line searching linked-list The station-line searching linked-list is also designed as a duplex linked-list, every of whose nodes stores all lines that run through the station. The list is sorted by station NO. to locates elements with binary search. Line-station search list structure is schematically shown in Fig. 4. Figure 4: Schematic view of station- line searching linked-list structure Similarly, the total time complexity of the station-line duplex searching linked-list is O(log(n)) + O(log(max(lps))) (5) where n is the count of all stations; max(lps) is the max count of lines that runs through a station.

52 46 Chunming Liu, Wansheng He, Hongming Xia and Zhongshe Gao No Implementation of algorithm The following route searching steps are acquired with integrated consider of search-level control strategy and optimization methods: With regard to the start station s start and the destination station s dest 1). Set L start and set L dest are obtained from station-line searching linked-list by binary searching and select the one with less elements, that is min(l start, L dest ). 2). Pick out all elements from min(l start, L dest ) one by one and search for s start or s dest in line-station searching linked-list (If min(l start, L dest ) = L start, search for s start, else search for s dest ), record matched lines and store them in set L match ; a). If L match, output all elements in L match which are direct lines; b). Else if L match =, deepen the search levels and continue with step 3); 3). Figure out D = A start A dest a). If D and D = {d 1, d 2,...d n }, then d 1, d 2,...d n are transfer stations. Go to step 1), perform direct-line searching with s start and elements in D, as well as elements in D and s dest respectively. b). Else if D = deepen the search levels and continue with step 4); 4). Pick out s start and s start from A start and A dest respectively, then set the searching level to 1 and continue with step 1), namely transform the problem to search for direct-line between s start and s dest. 4. Summary Because of the characteristics of intersecting and overlapping, the commonly used router algorithms based on Graph Theory or Genetic Algorithm are inapplicable in the public transport network; therefore, a set-based connectivity judging method was proposed in this article. Search level control strategy is adapted to resolve the graph problem with huge number of vertexes and edges and a hash table and two duplex linked-list are designed to improve the searching speed and reduce the searching complexity. References [1] Yuanzuo Li, Dishan Qiu, Properties of Generalized Connectedness of Graph and Commonly- Used Connecting Structures, Journal of Systems Engineering, 15(2000), No.3, 1-6. [2] Public Transportation Technology Forum of China, Beijing Public Transport Statistics, [3] Zhao Zhang, Fengxia Liu, Jixiang Meng, Super-Connected n-th Iterated Line Digraphs, Operations Research Transactions, 9(2005), No.6, [4] Zhao Zhang, Jixiang Meng, Super Edge Connectivity of Inifnite Circulants, Mathematica Applicata, 15(2002)(supplement), [5] Weimin Yan, Weimin Wu, Data Structure, Beijin, Tsinghua University Press, 1997,

53 Scientia Magna Vol. 4 (2008), No. 3, On the property of the Smarandache-Riemann zeta sequence Yanrong Xue Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract In this paper, some elementary methods are used to study the property of the Smarandache-Riemann zeta sequence and obtain a general result. Keywords Riemann zeta function, Smarandache-Riemann zeta sequence, positive integer. 1. Introduction and result For any complex number s, let ζ(s) = be the Riemann zeta function. For any positive integer n, let T n be a positive real number such that k=1 k s ζ(2n) = π2n T n, (1) where π is ratio of the circumference of a circle to its diameter. Then the sequence T = {T n } n=1 is called the Smarandache-Riemann zeta sequence. About the elementary properties of the Smarandache-Riemann zeta sequence, some scholars have studied it, and got some useful results. For example, in [2], Murthy believed that T n is a sequence of integers. Meanwhile, he proposed the following conjecture: Conjecture. No two terms of T n are relatively prime. In [3], Le Maohua proved some interesting results. That is, if ord(2, (2n)!) < 2n 2, where ord(2, (2n)!) denotes the order of prime 2 in (2n)!, then T n is not an integer, and finally he defies Murthy s conjecture. In reference [4], Li Jie proved that for any positive integer n 1, we have the identity ord(2, (2n)!) = α 2 (2n) + i=1 [ ] 2n 2 i = 2n a(2n, 2), where [x] denotes the greatest integer not exceeding x. So if 2n a(2n, 2) < 2n 2, or a(2n, 2) 3, then T n is not an integer.

54 48 Yanrong Xue No. 3 In fact, there exist infinite positive integers n such that a(2n, 2) 3, and T n is not an integer. From this, we know that Murthy s conjecture is not correct, because there exist infinite positive integers n such that T n is not an integer. In this paper, we use the elementary methods to study another property of the Smarandache- Riemann zeta sequence, and give a general result for it. That is, we shall prove the following conclusion: Theorem. If T n are positive integers, then 3 divides T n, more generally, if n = 2k, then 5 divides T n ; If n = 3k, then 7 divides T n, where k 0 is an integer. So from this Theorem we may immediately get the following Corollary. For any positive integers m and n(m n), if T m and T n are integers, then (T m, T n ) 3, (T 2m, T 2n ) 15, (T 3m, T 3n ) Proof of the theorem In this section, we shall complete the proof of our theorem. Lemmas which we state as follows: Lemma 1. If n is a positive integer, then we have First we need two simple where B 2n is the Bernoulli number. Proof. See reference [1]. Lemma 2. For any positive integer n, we have ζ(2n) = ( 1) n+1 (2π)2n B 2n, (2) 2(2n)! B 2n = I n p 1 2n 1 p, (3) where I n is an integer and the sum is over all primes p such that p 1 divides 2n. Proof. See reference [3]. Lemma 3. For any positive integer n, we have T n = (2n)!b n 2 2n 1 a n, (4) where a n and b n are coprime positive integers satisfying 2 b n, 3 b n, n 1. Proof. It is a fact that where ζ(2n) = ( 1) n 1 22n 1 π 2n B 2n, n 1, (5) (2n)! B 2n = ( 1) n 1 a n b n, n 1. (6) Using (1), (5) and (6), we get (4). Now we use above Lemmas to complete the proof of our theorem.

55 Vol. 4 On the property of the Smarandache-Riemann zeta sequence 49 For any positive integer n, from (4) we can directly obtain that if T n is an integer, then 3 divides T n, since (a n, b n ) = 1. From (1), (2) and (3) we have the following equality Let ζ(2n) = π2n = ( 1) n+1 (2π)2n B 2n n+1 (2π)2n = ( 1) T n 2(2n)! 2(2n)! p 1 2n p = p 1 p 2 p s, where p i (1 i s) is a prime number, and p 1 < p 2 < p s. Then from the above, we have I n p 1 2n 1, p T n = = = = ( 1) n+1 π 2n ( 1) n+1 (2n)! (2π) 2n 2(2n)! I n = 1 2 2n 1 I n 1 p p p 1 2n p 1 2n ( 1) n+1 (2n)! p 2 2n 1 I n p 1 2n p p 1 2n p 1 2n p p 1 2n 1 p ( 1) n+1 (2n)! p 1 p 2 p s ( 2 2n 1 I n p 1 p 2 p s p 1 p 2 p s ( 1 p p p s ( 1) n+1 (2n)! p 1 p 2 p s 2 2n 1 (I n p 1 p 2 p s p 2 p 3 p s p 1 p 3 p s p 1 p 2 p s 1 ) )) (7) Then we find that if p i p 1 p 2 p s, 1 i s, but p i (I n p 1 p 2 p s p 2 p 3 p s p 1 p 3 p s p 1 p 2 p s 1 ). So we can easily deduce that if T n are integers, when n = 2k, 5 can divide T n ; While n = 3k, then 7 can divide T n. This completes the proof of Theorem. References [1] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [2] A. Murthy, Some more conjectures on primes and divisors, Smarandache Notions Journal, 12(2001), 311. [3] Le Maohua, The Smarandache-Riemann zeta sequence, Smarandache Notions Journal, 14(2004), [4] Li Jie, On the Smarandache-Riemann zeta sequence, Research on Smarandache Problems in Number Theory, Hexis, Phoenix, AX, 2005,

56 Scientia Magna Vol. 4(2008), No. 3, The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 1 Baoan Chen and Chaofeng Shi Department of Mathematics, Xianyang Normal University, Xianyang, , P.R.China Abstract In this paper, a system of generalized set-valued variational inclusions in Banach spaces is introduced and studied, which includes many variational inclusions studied by others in recent years. By using some new and innovative techniques, several existence theorems for the system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces are established, and some perturbed iterative algorithms for solving this kind of system of set-valued variational inclusions are suggested and analyzed. generalize many known algorithms and results. Our results improve and Keywords System of generalized set-valued variational inclusion, iterative algorithm with e- rror, q-uniformly smooth Banach space 1. Introduction In recent years, variational inequalities have been extended and generalized in different directions, using novel and innovative techniques, both for their own sake and for the applications. Useful and important generalizations of variational inequalities are set-valued variational inclusions, which have been studied by [1-9]. Recently, in [1], Chaofeng Shi introduced and studied the following class of set-valued variational inclusion problems in a Banach space E. For a given m-accretive mapping A, B, C, G : E CB(E), nonlinear mappings N(, ), M(, ) : E E E, single-valued mapping g : H H, any given f E and λ > 0, find u E, x Au, ȳ Bu, z Cu, v Gu such that f N( x, ȳ) M(z, v) + λw(g(u)), (1.1) where CB(E) denotes the family of all nonempty closed and bounded subsets of E.Under the setting of q uniformly smooth Banach space, Chaofeng Shi [1] gave the existence and convergence theorem for the solution of variational inclusion (1.1). 1 This research is supported by the natural foundation of Shaanxi province of China(Grant. No.: 2006A14)and the natural foundation of Shaanxi educational department of China(Grant. No.:07JK421)

57 Vol. 4 The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 51 For a suitable choice of the mappings A, B, C, G, W, N, M, g and f E, a number of known and new variational inequalities, variational inclusions, and related optimization problems can be obtained from (1.1). On the other hand, X. P. Ding [2] introduced and studied the system of generalized vector quasi equilibrium problems in locally G-convex uniform spaces, and the existence results for the system of equilibrium problems is obtained. Inspired and motivated by the results in Chaofeng Shi [1], S. S. Chang [2], and X. P. Ding [3], the purpose of this paper is to introduce and study a class of system of set-valued variational inclusions. By using some new techniques, some existence theorems and approximate theorems for solving the set-valued variational inclusions in q-uniformly smooth Banach spaces are established and suggested. The results presented in this paper generalize, improve and unify the corresponding results of Chaofeng Shi [1], S. S. Chang [2], Ding [3,4], Huang [5,6], Zeng [7], Kazmi [8], Jung and Morales [9], Agarwal [10], Liu [11,12] and Osilike [13,14]. 2. Preliminaries Definition 2.1. Let I be a finite or infinite index set. Let {E i } i I be a family of real Banach spaces, A, B, C, G : E i CB(E i )be set-valued mappings, W i : D(W ) E i i I i I 2 Ei be an set-valued mapping, g i : E i E i be a single-valued mapping,and N i (, ), M i (, ) : E i E i E i be two nonlinear mappings, for any given f i E i and λ > 0, we consider the following problem of finding u E i, x Au, ȳ Bu, z Cu, v Gu such that, for each i I i I, f i N i (P i x, P i ȳ) M i (P i z, P i v) + λw i (g i (P i u)), (2.1) where P i : i I E i E i is a projection operator. This problem is called the generalized setvalued variational inclusion problem in a Banach space. Next we consider some special cases of problem (2.1). to (1) If I {i}, f i f, N i N, M i M, W i W, g i g, then problem (2.1) is equivalent f N( x, ȳ) M(z, v) + λw(g(u)), (2.2) This problem is introduced and studied by Chaofeng Shi [1]. (2) Furthermore, if M = 0, W = A : D(A) 2 E is an m-accretive mapping, A = T, B = F and C = G = 0, then problem(2.1) is equivalent to finding q E, w T q, v F q such that This problem was introduced and studied by S. S. Chang [1]. f N(w, v) + λa(g(q)). (2.3) For a suitable choice for the mappings A, B, C, G, W i, N i, M i, g i, f i and the family of spaces E i, we can obtain a lot of known and new variational inequalities, variational inclusions and the

58 52 Baoan Chen and Chaofeng Shi No. 3 related optimization problems. Furthermore, they can make us be able to study mathematics, physics and engineering science problems in a general and unified frame [1-9]. Definition 2.2. [10] Let E be a real Banach space. The module of smoothness of E is defined by ρ E (t) = sup{ 1 ( x + y + x y ) 1; x 1, y t}. 2 The space E is said to be uniformly smooth if lim(ρ E (t)/t) = 0. Moreover, E is called q- t 0 uniformly smooth if there exists a constant c > 0 such that ρ E (t) ct q. Remark 2.1. [10] All Hilbert spaces, L p (or l p ) spaces (p 2) and the Sobolev spaces, Wm(p p 2) are 2-uniformly smooth, while, for 1 < p 2, L p (or l p ) and Wm p spaces are p uniformly smooth. Definition 2.3. Let S : E CB(E) be a set-valued mapping. S is said to be quasicontractive, if there exists a constant r (0, 1) such that for any p Sx, q Sy, p q r max{ x y, x p, x q, y p, y q }. To prove the main result, we need the following lemmas. Lemma 2.1. [15] Let E be a q uniformly smooth Banach space with q > 1. Then there exists a constant c > 0 such that tx + (1 t)y z q [1 t(q 1)] y z q + tc x z q t(1 t q 1 c) x y q, for all x, y, z E and t [0, 1]. Lemma 2.2. [16] Suppose that {e n }, {f n }, {g n } and {γ n } are nonnegative real sequences such that e n+1 (1 f n )e n + f n g n + γ n, n 0, with {f n } [0, 1], n=0 f n =, lim n g n = 0 and n=0 γ n <. Then lim n e n = 0. For the remainder of this paper, r and c denote the constants appearing in definition 2.3 and lemma 2.1. We assume that H(, ) is the Hausdorff metric on CB(E) defined by H(A, B) = max{sup d(x, B), sup d(y, A)}, A, B CB(E). x A y B Lemma 2.3. [17] Let E be a complete metric space, T : E CB(E) be a set-valued mapping. Then for any given ε > 0 and any given x, y E, u T x, there exists v T y such that d(u, v) (1 + ε)h(t x, T y). Lemma 2.4. [18] Let X and Y be two Banach spaces, T : X 2 Y a lower semi-continuous mapping with nonempty closed and convex values. Then T admits a continuous selection, i. e., there exist a continuous mapping h : X Y such that h(x) T x, for each x X. Using Lemma 2.3 and Lemma 2.4, we suggest the following algorithms for the generalized set-valued variational inclusion (2.1).

59 Vol. 4 The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 53 Algorithm 2.1. For any given x 0 i I E i, x 0 Ax 0, y 0 Bx 0, z 0 Cx 0, v 0 Gx 0, compute the sequences {x n } and {y n } by the iterative schemes such that, for each i I, (1) P i x n+1 (1 α n )P i x n + α n (f i + P i y n N i (P i x n, P i ȳ n ) + M i (P i z n, P i v n ) λw i (g i (P i y n ))), (2) P i y n (1 β n )P i x n + β n (f i + P i x n N i (P i x n, P i y n) + M i (P i z n, P i v n) λw i (g i (P i x n ))), (3) x n Ay n, P i x n P i x n+1 (1 + 1 n + 1 )H i(p i Ay n, P i Ay n+1 ), (4) ȳ n By n, P i ȳ n P i ȳ n+1 (1 + 1 n + 1 )H i(p i By n, P i By n+1 ), (2.4) (5) z n Cy n, P i z n P i z n+1 (1 + 1 n + 1 )H i(p i Cy n, P i Cy n+1 ), (6) v n Gy n, P i v n P i v n+1 (1 + 1 n + 1 )H i(p i Gy n, P i Gy n+1 ), (7) x n Ax n, P i x n P i x n+1 (1 + 1 n + 1 )H i(p i Ax n, P i Ax n+1 ), (8) y n Bx n, P i y n P i y n+1 (1 + 1 n + 1 )H i(p i Bx n, P i Bx n+1 ), (9) z n Cx n, P i z n P i z n+1 (1 + 1 n + 1 )H i(p i Cx n, P i Cx n+1 ), (10) v n Gx n, P i v n P i v n+1 (1 + 1 n + 1 )H i(p i Gx n, P i Gx n+1 ), n = 0, 1, 2,. The sequence {x n } defined by (2.4), in the sequel, is called Ishikawa iterative sequence. In algorithm 2.1, if β n = 0, for all n 0, then y n = x n. Take x n = x n, ȳ n = y n, z n = z n and v n = v n, for all n 0 and we obtain the following. Algorithm 2.2. For any given x 0 i I E i, x 0 Ax 0, ȳ 0 Bx 0, z 0 Cx 0, v 0 Gx 0, compute the sequences {x n }, { x n }, {ȳ n }, {z n } and {v n } by the iterative schemes such that, for each i I, P i x n+1 (1 α n )P i x n + α n (f i + P i x n N i (P i x n, P i ȳ n ) + M i (P i z n, P i v n ) λw i (g i (P i x n ))), x n Ax n, P i x n P i x n+1 (1 + 1 n + 1 )H i(p i Ax n, P i Ax n+1 ), ȳ n Bx n, P i ȳ n P i ȳ n+1 (1 + 1 n + 1 )H i(p i Bx n, P i Bx n+1 ), (2.5) z n Cx n, P i z n P i z n+1 (1 + 1 n + 1 )H i(p i Cx n, P i Cx n+1 ), v n Gx n, P i v n P i v n+1 (1 + 1 n + 1 )H i(p i Gx n, P i Gx n+1 ), n = 0, 1, 2,. The sequence {x n } defined by (2.5), in the sequel, is called Mann iterative sequence.

60 54 Baoan Chen and Chaofeng Shi No An existence theorem for the system of the generalized set-valued variational inclusions In this section, we shall establish an existence theorem for solutions of the system of the set-valued variational inclusions (2.1). We have the following results. Theorem 3.1. Let E be a q-uniformly smooth Banach space, A, B, C, G : i I E i CB(E i ) be four set-valued mappings,for each i I, N i (, ), M i (, ) : E E E, g i : E i i I E i, W i : E i E i be four single-valued mappings. If I N i (P i A( ), P i B( ))+M i (P i C( ), P i G( )) λw i (g i (P i ( ))) is quasi contractive, then there exists u i I E i, x Au, ȳ Bu, v Cu, z Gu, which is a solution of system of the generalized set-valued variational inclusion (2.1). Proof. For each i I, let S(P i u) = P i u N i (P i A(u), P i B(u)) + M i (P i C(u), P i G(u)) λw i (g i (P i u)), S 1 (P i u) = f i + S(P i u). Since S is set-valued quasi contractive, S 1 is also set-valued quasi contractive. It follows from Naddler [17] that S 1 has a unique fixed point P i u E i, that is P i u S 1 (P i u) = f i + P i u N i (P i A(u), P i B(u)) + M i (P i C(u), P i G(u)) λw i (g(p i u)). Thus, there exist x Au, ȳ Bu, z Cu, v Gu such that, for each i I, f i N i (P i x, P i ȳ) M i (P i z, P i v) + λw i (g i (P i u)), that is, u E is a solution of the generalized set-valued variational inclusion (2.1). 4. Approximate problem of the system for generalized set-valued variational inclusions In theorem 3.1, under some conditions, we have proved that there exist u i I E i, x Au, ȳ Bu, z Cu, v Gu which is a solution of the system of generalized set-valued variational inclusion (2.1). In this section, we shall study the approximate problem for the solution of the system of generalized set-valued variational inclusions (2.1). We have the following result. Theorem 4.1. Let E i be a family of q-uniformly smooth Banach spaces, A, B, C, G : CB(E i ) be four set-valued continuous mappings, for each i I, E i i I i I N i (, ), M i (, ) : E i E i E i, g i : E i E i, w i : E i E i be four single-valued continuous mappings, and {α n }, {β n } be two sequences in [0,1] satisfying the following conditions: (i) I N i (P i A( ), P i B( )) + M i (P i C( ), P i G( )) λw i (g i P i ( )) is quasi contractive, (ii) for each i I, A, B, C, G are M-Lipschitz continuous with the constants µ 1, µ 2, µ 3 and µ 4,

61 Vol. 4 The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 55 (iii) 0 < h α n, n 0. 0 < µ i < 1/2, i = 1, 2, 3, 4, (iv) α n (q 1 cr q ) < 1, cr q < q 1, β q 1 n < 1/c(1 cr q ), cα q 1 n + cr q β n (cr q q + 1) 1 cr q. Then for any given x 0 i I E i, x 0 Ax 0, y 0 Bx 0, z 0 Cx 0, v 0 Gx 0, the sequences {x n }, { x n }, {ȳ n }, {z n } and {v n } defined by algorithm 2.1 strongly converge to the solution u i I E i, x Au, ȳ Bu, z Cu, v Gu of the system of generalized set-valued variational inclusions (2.1) which is given in Theorem 3.1, respectively. Proof. For each i I, in (1) and (2) of (2.4), choose h n W i (g i (P i x n )), k n W i (g i (P i y n )), such that P i x n+1 = (1 α n )P i x n + α n (f i + P i y n N i (P i x n, P i ȳ n ) + M i (P i z n, P i v n ) λk n ), P i y n = (1 β n )P i x n + β n (f i + P i x n N i (P i x n, P i y n) + M i (P i z n, P i v n) λh n ). Let p n = f i + P i y n N i (P i x n, P i ȳ n ) + M i (P i z n, P i v n ) λk n, r n = f i + P i x n N i (P i x n, P i y n) + M i (P i z n, P i v n) λh n. Then P i x n+1 = (1 α n )P i x n + α n p n, P i y n = (1 β n )P i x n + β n r n. (4.1) Since I N i (P i A( ), P i B( )) + M i (P i C( ), P i G( )) λw i (g i P i ( )) is quasi contractive, p n P i u r max{ P i y n P i u, p n P i y n }, which implies that p n P i u q r q ( P i y n P i u q + p n P i y n q ). (4.2) Note that p n r n rd(p i x n, P i y n ), n 0. We consider the following cases. Case 1: Suppose that d(p i x n, P i y n ) = P i x n P i y n for some n 0. It follows from (4.1) that p n r n r P i x n P i y n = r β n (P i x n r n ) rβ n P i x n r n. (4.3) Case 2: Suppose that d(p i x n, P i y n ) = P i x n r n for some n 0. Then we have p n r n r P i x n r n. (4.4)

62 56 Baoan Chen and Chaofeng Shi No. 3 Case 3: Suppose that d(p i x n, P i y n ) = P i y n r n for some n 0. Using (4.1), we have p n r n r P i y n r n = r (1 β n )(P i x n r n ) = r(1 β n ) P i x n r n r P i x n r n. (4.5) Case 4: Suppose that d(p i x n, P i y n ) = P i x n p n for some n 0. Then we have p n r n r P i x n p n. (4.6) Case 5: Suppose that d(p i x n, P i y n ) = P i y n p n for some n 0. It follows from (4.1) that which implies that It follows from (4.3)-(4.7) that It follows from Lemma 2.1 that p n r n r P i y n p n = r (1 β n )(P i x n p n ) + β n (r n p n ) r(1 β n ) P i x n p n + rβ n p n r n, p n r n r(1 β n) 1 rβ n P i x n p n r P i x n p n. (4.7) p n r n q r q P i x n p n + r q P i x n r n q, n 0. (4.8) P i y n P i u q = (1 β n )P i x n + β n r n P i u q = (1 β n )(P i x n P i u) + β n (r n P i u) q [1 β n (q 1)] P i x n P i u q + β n c r n P i u q Similarly, we have β n (1 β q 1 n c) P i x n r n q, n 0. (4.9) P i y n p n q [1 β n (q 1)] P i x n p n q + β n c p n r n q and β n (1 β q 1 n c) P i x n r n q, n 0. (4.10) P i x n+1 P i u q [1 α n (q 1)] P i x n P i u q + α n c p n P i u q α n (1 α q 1 n c) P i x n p n q, n 0. (4.11) By virtue of the condition (iii) and (iv), (4.2) and (4.8)-(4.11), we have P i x n+1 P i u q [1 α n (q 1)] P i x n P i u q

63 Vol. 4 The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 57 +α n cr q [ P i y n P i u q + P i y n p n q ] α n (1 αn q 1 c) P i x n p n q [1 α n (q 1) + cr q α n (1 β n (q 1))] P i x n P i u q +c 2 r q α n β n r n P i u q + c 2 r q α n β n p n r n q +[cr q α n (1 β n (q 1)) α n (1 αn q 1 c)] P i x n p n q 2cr q α n β n (1 βn q 1 c) P i x n r n q [1 α n (q 1) + cr q α n (1 β n (q 1)) + c 2 r 2q α n β n ] P i x n P i u q +2cr q α n β n (cr q + (βn q 1 c 1)) P i x n r n q +α n [c 2 r 2q β n + cr q (1 β n (q 1)) (1 αn q 1 c)] P i x n p n q [1 α n (q 1 cr q )(1 + cr q β n )] P i x n P i u q Set e n = P i x n P i u q, f n = α n (q 1 cr q ), [1 α n (q 1 cr q )] P i x n P i u q, g n = γ n = 0, n 0. It follows from the conditions (iii), (iv) and Lemma 2.2 that lim e n = 0, n that is, Note that which implies lim P ix n P i u = 0. n r n P i u r max{ P i x n P i u, r n P i x n }, r n P i u(n ). Thus, from (4.1), we have P i y n = (1 β n )P i x n + β n r n P i u. From (2.4) and conditions (ii) and (iii), we have P i x n P i x n+1 (1 + 1 n + 1 )H i(p i Ay n, P i Ay n+1 ) (1 + 1 n + 1 )µ P iy n P i y n+1, which implies that {P i x n } is a Cauchy sequence in E i. So, there exists x i I E i such that P i x n P i x. Now we show that x Au. In fact, d(p i x, P i Au) P i x P i x n + H i (P i Ax n, P i Au)

64 58 Baoan Chen and Chaofeng Shi No. 3 Also, we have P i x P i x n + µ(1 + 1 n + 1 ) P ix n P i u 0. ȳ n ȳ Bu, z n z Cu, v n v Gu. From (2.4), we have, for each i I, P i u f i + P i u N i (P i x, P i ȳ) + M i (P i z, P i v) λw i (g i (P i u)), that is, f i N i (P i x, P i ȳ) M i (P i z, P i v) + λw i (g i (P i u)). The required results. Remark 4.1. Theorem 4.1 generalizes Theorem 4.1 in Chaofeng Shi [1], the corresponding results in S. S. Chang [2], Agarwal [10], Liu [11, 12], Osilike [13, 14] and others. Remark 4.2. Since algorithm 2.2 is a special case of algorithm 2.1, from Theorem 4.1, we can obtain the convergence Theorem for algorithm 2.2, the details are omitted. References [1] Chaofeng Shi, Sanyang Liu, Generalized set-valued variational inclusions in q uniformly smooth Banach spaces, J. Math. Anal. Appl., 296(2004), [2] S. S. Chang, Set-valued variational inclusions in Banach spaces, J. Math. Anal. Appl., 248(2000), [3] X. P. Ding, Yao J. C. and Lin I. F., Solutions of system of system of generalized vector quasi-equilibrium problems in locally G-convex spaces, J. Math. Anal. Appl., 298(2004), [4] X. P. Ding, Iterative process with errors of nonlinear equation involving m-accretive operator, J. Math. Anal. Appl., 209 (1997), [5] N. J. Huang, Generalized nonlinear variational inclusions with noncompact valued mappings, Appl. Math. Lett., 9(1996), [6] N. J. Huang, On the generalized implicit quasi-variational inequalities, J. Math. Anal. Appl., 216(1997), [7] L. C. Zeng, Iterative algorithm for finding approximate solutions to completely generalized strongly nonlinear quasi-variational inequality, J. Math. Anal. Appl., 201(1996), [8] K. R. Kazimi, Mann and Ishikawa type perturbed iterative algorithms for generalized quasi-variational inclusions, J. Math. Anal. Appl., 209(1997), [9] J. S. Jung and C. H. Morales, The mann process for perturbed m-accretive operators in Banach spaces, submitted for publication. [10] R. P. Agarwal, Y. J. Cho, J. Li and N. J. Huang, Stability of iterative procedures with errors approximating common fixed points for a couple of quasi-contractive mappings in q-uniformly smooth Banach spaces, J. Math. Anal. Appl., 272 (2002),

65 Vol. 4 The system of generalized set-valued variational inclusions in q-uniformly smooth Banach spaces 59 [11] Q. H. Liu, On Naimpally and Singh s open questions, J. Math. Anal. Appl., 124(1987), [12] Q. H. Liu, A convergence theorem of the sequence of Ishikawa iteratives for quasicontractive mappings, J. Math. Anal. Appl., 146(1990), [13] M. O. Osilike, A stable iteration procedure for quasi-contractive mappings, Indian J. Pure Appl. Math., 27 (1996), [14] M. O. Osilike, Stability of the Ishikawa iteration method for quasi-contractive maps, Indian J. Pure Appl. Math., 28 (1997), [15] C. E. Chidume, M. O. Osilike, Fixed points iterations for quasi-contractive maps in uniformly smooth Banach spaces, Bull. Korean Math. Soc., 30 (1993), [16] L. S. Liu, Ishikawa and Mann iterative processes with errors for nonlinear strongly accretive mappings in Banach spaces, J. Math. Anal. Appl., 194(1995), [17] S. B. Naddler, Multivalued contraction mappings, Pacific J. Math., 30 (1969), [18] E. Michael, Continuous solutions, I. Ann. Math., 63(1956),

66 Scientia Magna Vol. 4(2008), No. 3, An iterative method for (R,S)-symmetric solution of matrix equation AXB = C Maolin Liang, Lifang Dai and Sanfu Wang College of Mathematics and Statistics, Tianshui Normal University, Tianshui, Gansu, , P. R. of China liangml2005@163.com Abstract For generalized reflexive matrices R, S, i.e., R T = R, R 2 = I, S T = S, S 2 = I, matrix A is said to be (R,S)-symmetric matrix when RAS = A. In this paper, an iterative method is established for solving the matrix equation AXB = C with (R,S)-symmetric solution X. By this method, the solvability of the above matrix equation can be determined autonomically, and a solution or the least-norm solution of which can be obtained within finite iteration steps. Meanwhile, the optimal approximation solution to a given matrix X 0 can also be derived by the least-norm solution of a new matrix equation A XB = C, where X = X X 0, C = C AX0B. Finally, given numerical examples illustrate the efficiency of the iterative method. Keywords Matrix equation, iterative method, (R,S)-symmetric solution, least-norm solution, optimal approximation solution. 1. Introduction Let R m n k be the set of all m n real matrices with rank k, SR n n, OR n n be the set of all symmetric matrices, orthogonal matrices in R n n, respectively. Denoted by the superscripts T and I n the transpose and identity matrix with order n, respectively. For matrices A = (a 1, a 2,..., a n ), B R m n, a i R m, R(A) and tr(a) represent its column space and trace, respectively; The symbol vec( ) represents the vec operator, i.e., vec(a) = (a T 1, a T 2,..., a T n ) T ; A B stands for the Kronecker product of matrices A and B; Moreover, A, B = tr(b T A) is defined as the inner product of the two matrices, which generates the Frobenius norm, i.e., A = < A, A > = tr(a T A). Definition 1.1. [1,2] For given generalized reflexive matrices R Rr m m, S R n n l, i.e., R T = R, R 2 = I, S T = S, S 2 = I, we say matrix A R m n is (R,S)-symmetric ((R,S)-skew symmetric), if RAS = A (RAS = A). The set of all m n (R,S)-symmetric ((R,S)-skew symmetric) matrices with respect to (R, S) is denoted by GSR m n (GSSR m n ). Definition 1.2. Assume M, N R p m, where p, m are arbitrary positive integers, if M, N = 0, matrices M, N are called to be orthogonal each other. 1 This work is supported by the Science Fundations of Tianshui Normal University (TSB0819 and TSB0721)

67 Vol. 4 An iterative method for (R,S)-symmetric solution of matrix equation AXB = C 61 Remark 1.1. (a) In this paper, we let R, S be generalized reflexive matrices as in Definition 1.1. (b) For F GSR m n, G GSSR m n, we can verify that F, G = 0. For above generalized reflexive matrix R, R T = R implies that the eigenvalues of R belong to real field, and the absolute of the eigenvalues equal to 1 since R 2 = I. Certainly, S has similar properties to R. Hence, we have the following assertion. Lemma 1.1. For given generalized reflexive matrices R, S as in Definition 1.1, there exist unitary matrices U 1 OR m m, and U 2 OR n n, such that R = U 1 I r 0 U1 T, S = U 2 I l 0 U2 T. 0 I m r 0 I n l From Definition 1.1 and Lemma 1.1, we generate the following conclusion. Lemma 1.2. Suppose X GSR m n, then X = U 1 where X 1 R r l, X 2 R (m r) (n l). X X 2 U T 2, From Lemma 1.2, we can easily get a (R,S)-symmetric matrix by choosing different A i, but U i (i = 1, 2) are fixed in Lemma 1.1. The problems to be discussed in this paper can be expressed as follows: Problem I. For given generalized reflexive matrices R R m m, S R n n, and A R p m, B R n q, C R p q, find X GSR m n, such that AXB = C. (1) Problem II. When Problem I is consistent, i.e., its solution set S E is not empty, for given matrix X 0 R m n, find ˆX S E such that ˆX X 0 = min X S E X X 0. (2) The linear matrix equation (1) has been discussed widely with unknown X [3 7], where X has particular structures, such as, symmetric, skew-symmetric, reflexive, anti-reflexive, Hermitian etc.. In these literatures, the authors obtained the expressions of associated solutions and solvability conditions by general inverse or matrix decompositions : the singular value decomposition (SVD), the generalized SVD (GSVD) [8] or the canonical correlation decomposition (CCD) [9]. In complex field, the inverse eigenvalue problem and procrustes problem on (R,S)-symmetric matrices have been studied by William [1,2]. Of course, the (R,S)-symmetric solution of matrix equation (1) can also be given by using GSVD or CCD. In addition, by establishing iterative algorithm, the symmetric solution, skew-symmetric solution, Hermitian least-norm solution of matrix equation (1) have been derived in [10-12], respectively. Moreover, the general solution and symmetric solution of matrix equations A 1 XB 1 = C 1, A 2 XB 2 = C 2 have also been investigated by Sheng [13] and Peng [14], respectively. Problem II is the optimal approximation problem, which occurs frequently in experimental design [13]. In equality (2), the matrix X 0 may be obtained from experiments, but it is not

68 62 Maolin Liang, Lifang Dai and Sanfu Wang No. 3 necessary to be needed form and the minimum residual requirement. The nearness matrix ˆX is the matrix that satisfies the needed form and the minimum residual restriction. Motivated by the iterative methods mentioned in [10], in this paper, we will construct an iterative method to obtain the (R,S)-symmetric solutions of Problem I and II. This work will be accomplished in section 2 and section 3, respectively. In section 4, we will provide some numerical examples to illustrate the efficiency of the iterative method. 2. The iterative method for Problem I In this section, we will propose an iterative algorithm to solve Problem I, some lemmas will be given to analyze the properties of this algorithm. For any (R,S)-symmetric matrix, We will show that the solution of matrix equation (1) can be obtained within finite iterative steps. The iterative method of Problem I is stated as follows: Algorithm 2.1. Step 1: Input matrices R R m m, S R n n, and A R p m, B R n q C R p q. Choosing arbitrary matrix X 1 GSR m n. Step 2: Calculate R 1 = C AX 1 B, P 1 = 1 2 (AT R 1 B T + RA T R 1 B T S), k := 1. Step 3: If R k = 0, stop; otherwise goto next step. Step 4: Calculate X k+1 = X k + R k 2 P k 2 P k. Step 5: Calculate R k+1 = C AX k+1 B= R k R k 2 P k 2 AP kb. P k+1 = 1 2 (AT R k+1 B T + RA T R k+1 B T S) + R k+1 2 R k 2 P k. Algorithm 2.1 reveals that X i, P i GSR m n (i = 1, 2,... ). In the sequel, we prove the feasibility of this algorithm. Lemma 2.1. The sequences R i, P i (i = 1, 2,... ) generated by Algorithm 2.1 satisfy that R i+1, R 1 = R i, R 1 R i 2 P i 2 P i, P 1, when j > 1, R i+1, R j = R i, R j R i 2 P i 2 P i, P j + R i 2 R j 2 P i 2 R j 1 2 P i, P j 1. (3) Proof. When j > 1, from the Algorithm 2.1, noting that RP i S = P i, we can obtain AP i B, R j = P i, A T R j B T = P i, AT R j B T + RA T R j B T S + P i, AT R j B T RA T R j B T S 2 2

69 Vol. 4 An iterative method for (R,S)-symmetric solution of matrix equation AXB = C 63 = P i, P j R j 2 R j 1 2 P j 1 + P i, AT R j B T RP i S, RAT R j B T S 2 2 = P i, P j R j 2 R j 1 2 P i, P j 1, (4) moreover, R i+1, R j = R i R i 2 P i 2 (AP ib), R j = R i, R j R i 2 P i 2 AP ib, R j = R i, R j R i 2 P i 2 P i, A T R j B T. (5) The proof of (5) implies that (3) holds when j = 1. Furthermore, submitting (4) into (5) generates (3) holds. The proof is completed. Lemma 2.2. The sequences {R i }, {P i } generated in the iterative process are self-orthogonal, respectively. i.e., R i, R j = 0, P i, P j = 0, i, j = 1, 2,..., k (k 2), i j. (6) Proof. We prove the conclusion by induction. Since F, G = G, F, we only proof (6) holds when i > j. When k = 2, it is clear from Lemma 2.1 that R 2, R 1 = R 1, R 1 R 1 2 P 1 2 P 1, P 1 = 0. Applying Algorithm 2.1 yields A T R 2 B T + RA T R 2 B T S P 2, P 1 = 2 = A T R 2 B T, P 1 + R 2 2 R 1 2 P 1, P 1 = R 2, AP 1 B + R 2 2 P 1 2 R R 2 2 R 1 2 P 1, P 1 = P 1 2 R 1 2 R 2, (R 1 R 2 ) + R 2 2 P 1 2 R 1 2 = 0. (7) Assume that (6) holds for k = s, that is, P s, P j = 0, P s, P j = 0, j = 1, 2,..., s 1. Being similar to the proof of (7), we can verify from Lemma 2.1 and the algorithm that R s+1, R s = 0, and P s+1, P s = 0. Now, we can finish the proof if R s+1, R j = 0, P s+1, P j = 0 hold. In fact, when j = 1, noting that the assumptions and (4), then R s+1, R 1 = R s, R 1 R s 2 P s 2 AP sb, R 1 = R s 2 P s 2 P s, A T R 1 B T

70 64 Maolin Liang, Lifang Dai and Sanfu Wang No. 3 = R s 2 P s 2 P s, P 1 = 0 and connecting with (3), we have A T R s+1 B T + RA T R s+1 B T S P s+1, P 1 = 2 A T R s+1 B T + RA T R s+1 B T S = 2 = A T R s+1 B T, P 1 = R s+1, AP 1 B = P 1 2 R 1 2 R s+1, (R 1 R 2 ) + R s+1 2 R s 2 P s, P 1, P 1 + R s+1 2 R s 2 P s, P 1 = P 1 2 R 1 2 R s+1, R 2 = 0. (8) Furthermore, when 2 j s 1, Lemma 2.1 and the assumptions imply that R s+1, R j = R s, R j R s 2 P s 2 P s, P j + R s 2 R j 2 P s 2 R j 1 2 P s, P j 1 = 0. Similar to the proof of (8), we can obtain P s+1, P j = 0. Hence, the conclusions hold. Remark 2.1. From Lemma 2.2, we know that, R i (i = 1, 2,..., pq) can be regarded as an orthogonal basis of matrix space R p q for their orthogonality each other, so a solution of Problem I can be obtained by the Algorithm 2.1 at most pq + 1 iteration steps. Lemma 2.3. When matrix equation (1) is consistent, provided that X is an arbitrary solution, then R k, P k generated by Algorithm 2.1 satisfy that X X k, P k = R k 2, k = 1, 2,.... Proof. When k=1, from the Algorithm 2.1 and Lemma 2.2, we get X X 1, P 1 = X X 1, AT R 1 B T + RA T R 1 B T S 2 = X, AT R 1 B T + RA T R 1 B T S X 1, AT R 1 B T + RA T R 1 B T S 2 2 = X, A T R 1 B T X 1, A T R 1 B T = A( X X 1 )B, R 1 = R 1 2. Assume that the conclusion holds for k = s, i.e., X X s, P s = R s 2, then X X s+1, P s = X X s R s 2 P s 2 P s, P s = ( X X s ), P s R s 2 = R s 2 R s 2 P s 2 R s 2 P s 2 P s, P s

71 Vol. 4 An iterative method for (R,S)-symmetric solution of matrix equation AXB = C 65 and X X s+1, P s+1 = = 0, X X s+1, AT R s+1 B T + RA T R s+1 B T S 2 = 1 2 X X s+1, A T R s+1 B T + RA T R s+1 B T S + R s+1 2 = X X s+1, A T R s+1 B T = A( X X s+1 )B, R s+1 = R s R s+1 2 R s 2 P s R s 2 X X s+1, P s Therefore, we complete the proof by the principle of induction. Remark 2.2. Lemma 2.3 implies that, if the linear matrix equation (1) is consistent, then, R i = 0 if and only if P i = 0. In other words, if there exist a positive number t such that R t 0 but P t = 0, then matrix equation (1) is inconsistent. Therefore, the solvability of Problem I can be determined automatically by Algorithm 2.1 in the absence of roundoff errors. Based on the previous analysis, we have the following conclusion whose proof is omitted. Theorem 2.1. When matrix equation (1) is consistent, for arbitrary initial matrix X 1 GSR m n, a solution of which can be obtained within finite iterative steps in the absence of roundoff errors. To facilitate the statement of our main results, we cite the following conclusion. Lemma 2.4. (see [10]) Suppose that the consistent linear equations My = b has a solution y 0 R(M T ), then y 0 is the least-norm solution of which. Theorem 2.2. When matrix equation (1) is consistent. If initial iteration matrix X 1 = A T HB T + RA T HB T S, where arbitrary H R p q, or especially, let X 1 = 0 R p q, then the solution generated by the iteration method is the unique least-norm solution of Problem I. Proof. Algorithm 2.1 and Theorem 2.1 imply that, if let X 1 = A T HB T + RA T HB T S, where H is an arbitrary matrix in R p q, a solution X of Problem I is of the form X = A T Y B T + RA T Y B T S. So it is enough to show that X is the least-norm solution of matrix equation (1). Considering the following linear matrix equations with X GSR m n AXB = C ARXSB = C (9) Obviously, the solvability of (9) is equivalent to that of matrix equation (1). Denote vec(x ) = x, vec(x) = x, vec(y ) = y, vec(c) = c, then the above matrix equations can be transformed into B T A x = c. (B T S) (AR) c In addition, the solution X can be rewritten as

72 66 Maolin Liang, Lifang Dai and Sanfu Wang No. 3 x = B T A (B T S) (AR) T y y R B T A (B T S) (AR) T, which implies, from Lemma 2.4, that X is the least-norm solution of the linear matrix equations (9). We complete the proof. 3. The solution of Problem II Suppose that matrix equation (1) is consistent, i.e., S E is not empty. It is easy to verify that S E is a closed convex set in matrix space GSR m n under Fronbenius norm, so the optimal approximation solution is unique. Without loss of generality, we can assume the given matrix X 0 GSR m n in Problem II because of the orthogonality between (R,S)-symmetric and (R,S)- skew symmetric matrix. In fact, for X S E, we have X X 0 2 = X X 0 + RX 0 S X 2 0 RX 0 S 2 2 = X X 0 + RX 0 S X 0 RX 0 S 2 Let X = X X 0, C = C AX0 B, then Problem II is equivalent to find the least-norm solution X GSR m n of the following matrix equation 2. A XB = C. (10) From Theorem 2.2, if let initial iteration matrix X 1 = A T HB T + P A T HB T P, where arbitrary H R p q, or especially, let X 1 = 0 R m n, we can obtain the unique least-norm solution X of matrix equation (10) by applying Algorithm 2.1. Furthermore, the unique optimal approximation solution ˆX to X 0 can be obtained by ˆX = X + X Numerical examples Example 1. Input matrices A, B, C, and generalized reflexive matrices R, S as follows: A = , R = ,

73 Vol. 4 An iterative method for (R,S)-symmetric solution of matrix equation AXB = C B =, S =, C = Next, we will find the associated solutions of Problems I and II by Algorithm 2.1 and Matlab software. For the influence of roundoff errors, R i will usually unequal to zero in the Fronbenius norm. Therefore, for any chosen positive number ε, however small enough, e.g., ε = 1.0e 010, whenever R k < ε, stop the iteration, and X k is regarded as a required solution. (I) The solutions of Problem I: Choose an arbitrary initial matrix X 1 = , by Algorithm 2.1 and 36 iteration steps, we obtain a solution of Problem I, X 36 = , with R 36 = e 011 < ε, and X 36 = If let initial matrix X 1 = A T HB T + P A T HB T P, where arbitrary H R 7 6, then the

74 68 Maolin Liang, Lifang Dai and Sanfu Wang No. 3 solution X k generated by Algorithm 2.1 is the unique least-norm solution of matrix equation (1) by Theorem 2.2. Choosing arbitrary matrix H = , then the least-norm solution X of Problem I is X = X 37 = , with R 37 = e 011 < ε, and X = (II) The solution of Problem II Suppose that the given matrix X 0 GSR 7 6 is X 0 = Compute C 0 = AX 0 B, let X = X X 0, C = C C0, then we can obtain the optimal approximation solution ˆX of Problem II by finding the least-norm solution X of equation (10), that is, X = , Hence the solution of Problem II is

75 Vol. 4 An iterative method for (R,S)-symmetric solution of matrix equation AXB = C ˆX = Example 2. Given matrices A, B, C being the same as example 2 in [8], A = , B = , C = , and R = , S = Choosing positive number ε = 1.0e 005, we will stop the iteration when R k < ε, or R k > ε, but P k < ε. Let initial iterative matrix X 1 = 0 R 4 4, by the iterative method, we obtain that R 6 = e > ε, and P 6 = e 009 < ε. Then, Theorem 2.1 implies that Problem I has no solution in GSR 4 4. References [1] F. William, Characterization and problems of (R,S)-symmetric, (R,S)-skew symmetric, and (R,S)-conjugate matrices, SIAM J. Matrix Anal. Appl., 26(2005), No.3, [2] F. William, Minimization problems for (R,S)-symmetric and (R,S)-skew symmetric matrices, Linear Algebra Appl., 389(2004), [3] K. Chu, Symmetric solutions of linear matrix equations by matrix decompositions. Linear Algebra Appl., 119(1989), [4] H. Dai, On the symmetric solutions of linear matrix equations, Linear Algebra Appl., 131(1990), 1-7. [5] F. Zhou, Solutions of matrix equation AXA T = C in symmetric and skew-symmetric matrix set (In Chinese), J. Math. Research Exposition, 25(2005), No.2, [6] X. Peng, X. Hu, L. Zhang, The reflexive and anti-reflexive solutions of the matrix equation A XB = C, J. Comput. Appl. Math., 200(2007), No.2, [7] M. Wang, M. Wei, On weighted least-squares Hermitian solution of matrix equation AXB = E (In Chinese), Math. Numer. Sinica, 26(2004), No.2, [8] K. Chu, Singular value and generalized singular value decompositions and the solution of linear matrix equations, Linear Algebra Appl., 88(1987),

76 70 Maolin Liang, Lifang Dai and Sanfu Wang No. 3 [9] G. Golub, H. Zha, Perturbation analysis of the canonical correlations of matrix pairs, Linear Algebra Appl., 210 (1994), [10] Y. Peng, X. Hu, L. Zhang, An iterative method for the symmetric solutions and the optimal approximation solution of the matrix equation AXB = C, Appl. Math. Comput., 160(2005), [11] G. Huang, F. Yin, K. Guo, An iterative method for the skew-symmetric solution and the optimal approximate solution of the matrix equation AXB = C, J. Comput. Appl. Math., 212(2008), No.2, [12] Y. Deng, Z. Bai, Y. Gao, Iterative orthogonal direction methods for Hermitian minimum norm solutions of two consistent matrix equations, Linear Algebra Appl., 13(2006), [13] X. Sheng, G. Chen, A finite iterative method for solving a pair of linear matrix equations (AXB,CXD)=(E,F), Appl. Math. Comput., 189(2007), No.2, [14] Y. Peng, X. Hu, L. Zhang, An iteration method for symmetric solutions and optimal approximation solution of the system of matrix equations A 1 XB 1 = C 1, A 2 XB 2 = C 2, Appl. Math. Comput., 183(2006), No.2, [15] T. Meng, Experimental design and decision support, in Leondes(Ed.), Expert Systems, the Technology of Knowledge Management and Decision Making for the 21th Century, Academic Press, 1(2001).

77 Scientia Magna Vol. 4 (2008), No. 3, On an equation of the Smarandache function Lixiang Cai Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any positive integer n, let S(n) denotes the Smarandache function. The main purpose of this paper is using the elementary method to study the solvability of the equation S(a) 2 + S(b) 2 = S(c) 2, and obtain its all positive integer solutions. Keywords The Smarandache function, Pythagorean triple, equation, positive integer solutions. 1. Introduction and result For any positive integer n, the famous Smarandache function S(n) is defined as the smallest positive integer m such that m! is divisible by n. That is, S(n) = min{m : n m!, m N}, where N denotes the set of all positive integers. For example, the first few values of S(n) are: S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 4, S(5) = 5, S(6) = 3, S(7) = 7, S(8) = 4, S(9) = 6, S(10) = 15, S(11) = 11, S(12) = 4, S(13) = 13, S(14) = 7, S(15) = 5, S(16) = 6, S(17) = 17, S(18) = 6, S(19) = 19, S(20) = 5,. From the definition of S(n) we can easily get the following conclusions: if n = p α, S(n) = αp, α p; S(n) < αp, α > p. Let n = p α1 1 pα2 2 pαr r denotes the factorization of n into prime powers, then S(n) = max 1 i r {S(pαi i )}. About other properties of S(n), many scholars had studied it, and obtained some interesting results, see [1], [2], [3], [4] and [5]. For example, in reference [5], Xu Zhefeng proved that for any real number x > 1, (S(n) P (n)) 2 = 2ζ( 3 2 )x ( ) ln x + O x 3 2 ln 2, x n x where P (n) is the greatest prime divisor of n, ζ(s) is the Riemann zeta-function. In reference [6], Rongi Chen and Maohua Le proved that the equation S 2 (n) + S(n) = kn has infinitely many positive integer solutions for every positive integer k. On the other hand, Bencze [7] proposed the following problem: Find all positive integer solutions of the equation 1 S(a) S(b) 2 = 1 S(c) 2. (1)

78 72 Lixiang Cai No. 3 About this problem, it seems that none had studied it yet, at least we have not seen any related papers before. In this paper, we use the elementary method to study this problem, and solved it completely. That is, we shall prove the following: Theorem. The equation (1) has infinite positive integer solutions. And each (a, b, c) satisfy the equation (1) if and only if S 2 (a) + S 2 (b) = z 2, which means S(a), S(b), z are Pythagorean triples, and GCD(S(a), S(b)) = d > 1, z d 2, S(c) = xy, where GCD(S(a), S(b)) z denotes the greatest common divisor of S(a) and S(b). From this Theorem we can obtain many positive integer solutions of the equation (1). For example, if we take S(a) d = 3, S(b) d = 4, z d = 5, d = 5, S(c) = 12, then we can get a = 53, b = 5 4, c = 3 5. If d = 10, then a = 5 7, b = 5 9, c = Proof of the theorem In this section, we shall complete the proof of our theorem. First we need the following: Lemma. For every positive integer m, the equation S(n) = m has positive integer solution. Proof. From the definition of S(n) and reference [8] we can easily get this conclusion. Now we use this Lemma to prove our theorem. For convenience, we denote S(a) by x, S(b) by y, S(c) by w. Then the equation (1) become 1 x y 2 = 1 w 2. (2) It is clear that if one of S(a), s(b) and S(c) is 1, then (a, b, ) does not satisfy the equation (1). So without loss of generality we can assume that x 2, y 2, w 2. Obviously the solvability of the equation (1) is equivalent to the solvability of the equation (2). Now we consider the positive integer solutions of the equation (2). From equation (2) we can obtain and x 2 y 2 = (x 2 + y 2 )w 2 (3) x 2 y 2 x 2 + y 2 = w2. (4) Suppose that the equation (1) has positive integer solutions, then there exist x, y, w satisfied the equation (3). Noting that x 2, y 2, w 2, form the equation (3) we can get that x 2 + y 2 is a complete square, in other words, there exists an positive integer z such that x 2 + y 2 = z 2, and so x, y, z are Pythagorean triples. Hence we can get ( xy ) 2 = w 2. (5) z Suppose GCD(x, y) = d, we can get GCD(x, y) = GCD(x, z) = GCD(y, z) = d in equation (5). Suppose x = dx, y = dy, z = dz, then we have ( dx y ) 2 = w 2, (6) z

79 Vol. 4 On an equation of the Smarandache function 73 from equation (6), w is integer, then z dx y. Note that GCD(x, z ) = GCD(y, z ) = 1 and GCD(x y, z ) = 1, then we can get z d and d z > 2. Because z = dz, z d 2, we can get w = xy from equation (5). z At last, we get the positive integer solutions of equation (2), and the solutions satisfying x 2 +y 2 is a complete square, in other words, there exists a positive integer z such that x 2 +y 2 = z 2, and GCD(x, y) = d > 2, z d 2, w = xy. From Lemma we can get the positive integer z solutions of equation (2). Since there are infinite Pythagorean triples x, y, z satisfied the above situation, so the equation (2) has infinite positive integer solutions. And the solvability of equation (1) is equivalent to the solvability equation (2), so the equation (1) also has infinite positive integer solutions. This completes the proof of Theorem. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] I. Balacenoiu and V. Seleacu, History of the Smarandache function, Smarandache Notions Journal, 10(1999), No.1-2-3, [3] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, [4] Lu Yaming, On the solutions of an equation involving the Smarandache function, Scientia Magna, 2(2006), No.1, [5] Xu Zhefeng, On the value distribution of the Smarandache function, Acta Mathematica Sincia (in Chinese), 49(2006), No.5, [6] Rongi Chen and Maohua Le, On the fuanctional equation S 2 (n) + S(n) = kn, Smarandache Notions Journal, 11(2000), No.1-2-3, [7] M. Bencze, Open questions for the Smarandache function, Smarandache Notions Journal, 12(2001), No.1-2-3, [8] Charles Ashbacher, An Introduction to the Smarandache Function, Erhus University Press, Vail, 1995.

80 Scientia Magna Vol.4 (2008), No. 3, On the Smarandache divisibility theorem Chunping Wang Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract In this paper, we used the elementary method to study the divisibility properties of the integer, and improved the Smarandache divisibility theorem. Keywords Divisibility theorem, Gauss function. 1. Introduction and result Divisibility theory is the fundamental part in mathematics. The study of the divisibility theory is very important and interesting contents in number theory, many authors pay much attention to this topic, and some deeply results have been obtained. For example, F.Smarandache [1] proved that for any positive integer m, we have m (a m a) (m 1)!. Maohua Le [3] improved the above result, and obtained the following conclusion: [ m ] m (a m a)! 2 Because (m, m 1) = 1, it is obvious that m (a m a) (m 2)!. In this paper, we using the elementary method to study this problem, and give another form of the Smarandache divisibility theorem. That is, we shall prove the following conclusion: Theorem. If a and m are two integers with m > 1, then we have m (am a) (m 2)! [ m ], where [x] denotes the largest integer less than or equal to x. 2. Some preliminary lemmas In this section, we shall give several simple Lemmas which are necessary in the proof of our theorem. They are stated as follows: Lemma 1. If a and m are two integers with m > 0, then (a m a)(m 2)! can be divided by m. Proof. (See reference [2]). Lemma 2. Let a and b are two positive integers, S = [ {1, 2, 3,..., a}. Then the number a ] of all positive integers in S which can be divided by b is. b

81 Vol. 4 On the Smarandache divisibility theorem 75 Proof. (See reference [1]). Lemma 3. Little Fermat s Theorem: For any integer a and any prime p, we have the congruence Proof. (See reference [3]). a p a mod p. 3. Proof of the theorems In this section, we shall complete the proof of our theorem directly. It is clear that we just need to prove that 4 [ m 1 Now we separate m into three cases: 4 ] (am a) (m 2)!. (1) m (a). If m = 1, 2, 3, 4, then from Lemma 1 we know that (1) is true. [ ] [ ] m 1 m 1 (b). If m = 5, then = = 4. 4 (a m a) (m 2)! m = (a5 a)3! 5 = (a 1)a(a + 1)(a2 + 1)3!. 5 We also know that there exists at least an even number among a, (a + 1) and (a 1). So we must have 4 (a 5 a) (5 2)!, note that (4, 5) = 1, we may immediately deduce that (1) is correct. [ ] [ ] m 1 m 1 If m = 8, then = = 4. 4 (a m a) (m 2)! m = (a8 a) 6! 8 = a (a7 1) 6!. 8 It is clear that there exists an even number between a and (a 7 1), so we have 2 (a 8 a), 4 2 6!. That is to say, our Theorem is true if m = 8. (c). If m > 5, and m 8, then we can discuss it in two cases: If m is an odd number, we can write m = 4k + t, where t = 1 or 3, k N. Then 4 [ m 1 If m = 4k + 1, then we have the estimate [ ] m 2 j=1 2 j = 4 ] [ m 2 4k + t 1 4 = 4 2 ] + [ m 2 4 = 4 k. ] + = (2k 1) + (k 1) + (3k 2) 2k. (2)

82 76 Chunping Wang No. 3 If m = 4k + 3, then we also have [ ] m 2 2 j j=1 = [ m 2 2 ] + [ m 2 4 ] + = 2k + k + 3k > 2k. (3) From (2) and (3) we may immediately deduce that 4 k is a divisor of (4k 2)!. Let (m 2)! = 4 k D, where D is a positive integer. Note that (4k + t, 4 k ) = 1, so (4k + t) (a m a) D. Therefore, if m > 5 is an odd number, then (1) is correct. If m is an even number, we can write m = 2 r q, where q is an odd number. It is clear that if r = 1, then from the above we know that the (1) is true. So we can assume that r 2. This time, in order to prove our theorem, we just need to prove that 2 2(r 1) q 2+r (a2rq a)(2 r q 2)!. q Similarly, we also have [ ] m 2 2 j j=1 = [ m 2 2 ] + [ m 2 4 ] + + [ m 2 2 r ] + = 2 r 1 q r 2 q q r 1 q r 2 q q 1 = q(2 r r ) r = q2 r q r. (4) Note that r 2 and (2, q) = 1, so we have (q 2 r q r) (2 (r 1) q 2+r) = 2 r 1 q q+2 2r 0. Thus, 2 2r 1 q 2+r is an divisor of (m 2)!. Since q m 2, we also have q (m 2)!. Therefore, from (2), (3) and (4) we know that (1) is correct, if m > 5 ( m 8 ) is an even number. Combining (a), (b) and (c) we may immediately complete the proof of our Theorem. References [1] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [2] F.Smarandache, Problems avec et sans...problems!, Somipress, Fes, Morocco, 1983, Problem 7, 140, [3] Maohua Le, An improvement on the Smarandache divisibility theorem, Smarandache Notions Journal, 10(1999), No.1-2-3, [4] Pan Chengdong, Pan Chengbiao, The elementary number theory (in Chinese), Beijing University Press, Beijing, [5] Florentin Smarandache, Sequences of numbers involved in unsolved problems, Hexis, America, 2006.

83 Scientia Magna Vol.4 (2008), No. 3, Some properties of the permutation sequence Min Fang Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract The main purpose of this paper is using the elementary method to study the properties of the permutation sequence, and prove some interesting conclusions. Keywords Permutation sequence, perfect power, divisibility. 1. Introduction and result For any positive integer n, the F.Smarandache permutation sequence {P (n)} is defined as P (n) = 135 (2n 1)(2n) 42. For example, the first few value of the sequence {P (n)} are: P (1) = 12, P (2) = 1342, P (3) = , P (4) = ,. This sequence was introduced by professor F.Smarandache in reference [1], where he asked us to study its elementary properties. About this problem, many people had studied it, and obtained a series valuable results, see references [2], [3] and [4]. In reference [5], F.Smarandache proposed the following problem: Is there any perfect power among the permutation sequences? That is, whether there exist positive integers n, m and k with k 2 such that P (n) = m k. The main purpose of the paper is using the elementary method to study this problem, and solved it completely. At the same time, we also obtained some other properties of the permutation sequence {P (n)}. That is, we shall prove the following: Theorem 1. There is no any perfect power among the permutation sequence. Theorem 2. Among the permutation sequence {P (n)}, there do not exist the number which have the factor 2 k, where k 2; There exist infinite positive integers a {P (n)} such that 3 2 divide a. 2. Proof of the theorems In this section, we shall use the elementary method to prove our Theorems directly. First we prove Theorem 1. For any positive integer n 2, note that P (n) be an even number and P (n) = 135 (2n 1)(2n) 42 = 10 α α2 + + (2n 1) 10 αn + (2n) 10 αn (1) So from (1) we know that P (n) = 10 α α2 + + (2n 1) 10 αn + (2n) 10 αn mod 2

84 78 Min Fang No. 3 and P (n) = 10 α α2 + + (2n 1) 10 αn + (2n) 10 αn mod 4. (2) That is to say, 2 P (n) and 4 P (n). So P (n) is not a perfect power. Otherwise, we can write P (n) = m k, where k and m 2. Since P (n) be an even number, so m must be an even number. Therefore, P (n) = m k 0 mod 4. Contradiction with 4 P (n). This proves Theorem 1. Now we prove Theorem 2. For any positive integer n 2, from (2) we may immediately deduce that 4 r P (n), where r 2 is an integer. Now we write P (n) as: P (n) = 10 α α2 + + (2n 1) 10 αn + (2n) 10 αn (3) Note that 10 1 mod 9, from the properties of the congruence we may get 10 t 1 mod 9 for all integer t 1. Therefore, P (n) = 10 α α2 + + (2n 1) 10 αn + (2n) 10 αn n n (2n + 1) mod 9. (4) From (4) we know that P (n) 0 mod 9, if n 0 or 4 mod 9. That is to say, for all integers n 2, P (9n) and P (9n + 4) can be divided by 9 = 3 2. This completes the proof of Theorems. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 21, [2] Tom M.Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [3] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [4] P.K.Guy, Unsolved Problems in Number Theory, Springer-Verlag, New York, [5] Smarandache Sequences at [6] Smarandache Sequences at [7] Smarandache Sequences at

85 Scientia Magna Vol.4 (2008), No. 3, One problem related to the Smarandache quotients Xiaojun Qi Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Department Of Mathematics, Tianshui Normal University, P.R.China Abstract For any positive integer n 1, the Smarandache quotients Q(n) is defined as the smallest positive integer k such that n k is a factorial number. That is, Q(n) = min{k : n k = m!}. The main purpose of this paper is using the elementary method to study the properties of the Smarandache quotients sequence, and give an identity involving the Smarandache quotients sequence. Keywords The Smarandache quotients, identity, infinite series, elementary method. 1. Introduction and results In his book Only problems, not solutions, professor F.Smarandache introduced many functions, sequences and unsolved problems, many authors had studied it, see references [1], [2] and [3]. One of the unsolved problems is the Smarandache quotients sequence {Q(n)}, it is defined as the smallest positive integer k such that n k is a factorial number. That is, Q(n) = min{k : n k = m!}, where m is a positive integer. For example, from the definition of Q(n) we can find that the first few values of Q(n) are Q(1) = 1, Q(2) = 1, Q(3) = 2, Q(4) = 6, Q(5) = 24, Q(6) = 1, Q(7) = 720, Q(8) = 3, Q(9) = 80, Q(10) = 12, Q(11) = , Q(12) = 2, Q(13) = , Q(14) = 360, Q(15) = 8, Q(16) = 45,. In reference [4], professor F.Smarandache asked us to study the properties of the sequence {Q(n)}. About this problem, some authors had studied it, and obtained several simple results. For example, Kenichiro Kashihara [5] proved that for any prime p, we have Q(p) = (p 1)!. In this paper, we use the elementary method to study the properties of an infinite series involving the Smarandache quotients sequence, and give an interesting identity. That is, we shall prove the following conclusion: Theorem 1. Let d(n) denotes the Dirichlet divisor function, then we have the identity + n=1 1 + Q(n) n = m=1 m d(m!) (m + 1)!. Let Q 1 (2n 1) denotes the smallest positive odd number such that Q 1 (2n 1) (2n 1) is a two factorial number. That is, Q 1 (2n 1) (2n 1) = (2m 1)!!, where (2m 1)!! = (2m 1). Then for the sequence {Q 1 (2n 1)}, we can also get the following:

86 80 Xiaojun Qi No. 3 Theorem 2. Let d(n) denote the Dirichlet divisor function, then we also have the identity + n=1 1 + Q 1 (2n 1) (2n 1) = 2 2. Proof of the theorems m=1 m d((2m 1)!!). (2m + 1)!! In this section, we shall use the elementary method to complete the proof of our theorems directly. For any positive integer n, let Q(n) = k, from the definition of Q(n) we know that k is the smallest positive integer such that n k = m!. So k must be a divisor of m!. This implies that k m!, and the number of all k (such that k m!) is d(m!). On the other hand, if k (m 1)!, then Q(n) k, and the number of all k (such that k (m 1)!) is d((m 1)!). So the number of all k (such that k m! and k (m 1)!) is d(m!) d((m 1)!). That is means, the number of all k (such that Q(n) = k and Q(n) n = m! ) is d(m!) d((m 1)!). From this we may immediately get + n=1 1 + Q(n) n = + m=1 n=1 Q(n) n=m! + = 1 + = m=2 + m=1 1 + m! = m=1 d(m!) d((m 1)!) m! (m + 1) d(m!) d(m!) (m + 1)! #{Q(n) n = m!} m! = = + m=1 + m=1 d(m!) m! m d(m!) (m + 1)!, + m=1 d(m!) (m + 1)! where d(m) is the Dirichlet divisor function, and #{Q(n) n = m!} denotes the number of all solutions of the equation Q(n) n = m!. This proves Theorem 1. Similarly, we can also deduce Theorem 2. This completes the proof of Theorems. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [3] Tom M.Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [4] F. Smarandache, Sequences of numbers involved in unsolved problems, Hexis, Phoenix, [5] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, 1996.

87 Scientia Magna Vol. 4 (2008), No. 3, The Smarandache bisymmetric determinant natural sequence A.A.K. Majumdar Ritsumeikan Asia-Pacific University, APU, 1-1 Jumonjibaru, Beppu-shi , Oita-ken, Japan majumdar@apu.ac.jp/aakmajumdar@gmail.com Abstract Murthy [1] introduced the Smarandache Bisymmetric Determinant Natural Sequence. In this paper, we derive the sum of the first n terms of the sequence. Keywords The Smarandache bisymmetric determinant natural sequence, the n-th term, the sum of the first n terms. 1. Introduction The Smarandache bisymmetric determinant natural sequence (SBDNS), introduced by Murthy [1], is defined as follows. Definition 1.1. The Smarandache bisymmetric determinant natural sequence, {SBDNS(n)}, is 1, , , , A first few terms of the sequence are 1, -3, -8, 20, 48, -112, -256, 576,. The following result is due to Majumdar [2]. Theorem 1.1. Let a n be the n-th term of the Smarandache bisymmetric determinant natural sequence. Then, n 2 n 1 n n 1 n n 1 [ n ] a n = n n 1 n 2 = ( 1) 2 (n + 1) 2 n 2.. n 2 n 1 n n 1 n n n n 1 n

88 82 A.A.K. Majumdar No. 3 Let {S n } be the sequence of n th partial sums of the sequence {a n }, so that S n = n a k, n 1. k = 1 This paper gives explicit expressions for the sequence {S n }. This is given in Theorem 3.1 in Section 3. In Section 2, we give some preliminary results that would be necessary for the proof of the theorem. We conclude this paper with some remarks in the final section, Section Some preliminary results In this section, we derive some preliminary results that would be necessary in deriving the expressions of S n in the next section. These are given in the following two lemmas. Lemma 2.1. For any integer m 1, k=1,3,...,(2m 1) 2 2(k 1) = 1 15 (24m 1). Proof. Since k=1,3,...,(2m 1) 2 2(k 1) = (m 1) is a geometric series with common ratio 2 4, the result follows. Lemma 2.2. For any integer m 1, k=1,3,...,(2m 1) k 2 2(k 1) = 1 15 (2m 1)24m (24m+1 17). Proof. Denoting by s the series on the left above, we see that s = (2m 1) 2 4(m 1), ( ) so that, multiplying throughout by 2 4, we get 2 4 s = (2m 3) 2 4(m 1) + (2m 1) 2 4m ( ) Now, subtracting (**) from (*), we have [ (1 2 4 )s = (m 2)] (2m 1) 2 4m { 2 = (m 1) } (2m 1) 2 4m 1 = 1 ( 2 4m+1 17 ) (2m 1) 2 4m 15 which now gives the desired result.

89 Vol. 4 The Smarandache bisymmetric determinant natural sequence Main results In this section, we derive the explicit expressions of the n-th partial sums, S n, of the Smarandache bisymmetric determinant natural sequence. From Theorem 1.1, we see that, for any integer k 1, a 2k + a 2k+1 = ( 1) k (6k + 5)2 2(k 1), a 2k+2 + a 2k+3 = ( 1) k+1 (6k + 11)2 2k, so that a 2k + a 2k+1 + a 2k+2 + a 2k+3 = 3( 1) k+1 (6k + 13)2 2(k 1). (1) Letting S n = a 1 + a a n, we can prove the following result. Theorem 3.1. For any integer m 0, 1). S 4m+1 = 3 5 m 22(2m+1) m 6 25 = 2 { (60m + 31)2 4m 1 3 } ; 25 2). S 4m+2 = 1 5 m 24m (2m+1) 6 25 = ). S 4m+3 = 3 5 m 24(m+1) (2m+1) 6 25 = ). S 4m+4 = 1 5 m 24m (2m+2) 6 25 = 2 25 { (10m + 11)2 4m } ; { (60m + 61)2 4m } ; { (5m + 8)2 4(m+1) 3 Proof. To prove the theorem, we make use of Lemma 2.1 and Lemma 2.2, as well as Theorem ). Since S 4m+1 can be written as S 4m+1 = a 1 + a a 4m+1 = a 1 + by virtue of (1), S 4m+1 = a = a k=1,3,...,(2m 1) k=1,3,...,(2m 1) k=1,3,...,(2m 1) }. (a 2k + a 2k+1 + a 2k+2 + a 2k+3 ), ( 1) k+1 (6k + 13)2 2(k 1) k2 2(k 1) + 13 k=1,3,...,(2m 1) 2 2(k 1) Now, appealing to Lemma 2.1 and Lemma 2.2, we get [ 6 S 4m+1 = (2m 1)24m 2 ] 25 (24m+1 17) (24m 1).

90 84 A.A.K. Majumdar No. 3 which now gives the desired result after some algebraic manipulations. 2). Since S 4m+2 = S 4m+1 + a 4m+2, from part (1) above, together with Theorem 1.1, we get [ 3 S 4m+2 = 5 m 22(2m+1) m 6 ] (4m + 3) 2 4m, 25 which gives the desired expression for S 4m+2 after algebraic simplifications. 3). Since S 4m+3 = S 4m+2 + a 4m+3 = we get the desired expression for S 4m+3 after simplifications. 4). Since S 4m+4 = S 4m+3 + a 4m+4 = the result follows after some algebraic simplifications. The case when m = 0 can easily be verified. Hence, the proof is complete. [ 1 5 m 24m (2m+1) 6 ] (4m + 4) 2 4m+1, 25 [ 6 5 m 24m (2m+1) 6 ] (4m + 5) 2 4m+2, Remarks Theorem 3.1 in the previous section gives the n-th term of the sequence of partial sums, {S n }, in all the possible four cases. The following lemmas prove that, in each case, S n is indeed an integer. Lemma 4.1. For any integer m 0, 2 4m 1 (60m + 31) 3 is divisible by 25. Proof. The result is true for m = 0, 1. So, we assume its validity for some positive integer m. Now, since [2 4m+3 {60(m + 1) + 31} 3] {2 4m 1 (60m + 31) 3} = 25(36m + 57)2 4m 1, it follows, by virtue of the induction hypothesis, that 2 4m+3 {60(m+1)+31} 3 is also divisible by 25. Thus, the result is true for m + 1 as well, completing induction. Lemma 4.2. For any integer m 0, 2 4m+1 (10m + 11) + 3 is divisible by 25. Proof. is by induction on m. The result is clearly true for m = 0, 1. Now, assuming its validity for some positive integer m, since [2 4m+5 {10(m + 1) + 11} + 3] {2 4m+1 (10m + 11) + 3} = 25(6m + 13)2 4m+1, is divisible by 25, it follows that the result is true for m + 1 as well. This completes the proof. Lemma 4.3. For any integer m 0, 2 4m+1 (60m + 61) + 3 is divisible by 25.

91 Vol. 4 The Smarandache bisymmetric determinant natural sequence 85 Proof. is by induction on m. The result is clearly true for m = 0, 1. We now assume that the result is true for some positive integer m. Then, since [2 4m+5 {60(m + 1) + 61} + 3] {2 4m+1 (60m + 61) + 3} = 25(36m + 75)2 4m+1, this, together with the induction hypothesis, shows that the result is true for m + 1 as well. This completes the proof by induction. Lemma 4.4. For any integer m 0, 2 4(m+1) (5m + 32) 3 is divisible by 25. Proof. We first assume that the result is true for some positive integer m. Now, since [2 4(m+2) {5(m + 1) + 13} 3] {2 4(m+1) (5m + 13) 3} = 25(3m + 8)2 4(m+1), this together with the induction hypothesis, shows that 2 4(m+2) {5(m + 1) + 13} 3 is also divisible by 25. This, in turn, shows that the result is true for m + 1 as well. To complete, we have to prove the validity of the result for m = 0, 1, which can easily be checked. The Smarandache bisymmetric arithmetic determinant sequence, introduced by Murthy [1], is a a + d a + 2d a a + d a, a + d a, a + d a + 2d a + d,.... a + 2d a + d a The n-th term of the above sequence has been found by Majumdar [2] to be [ n ] ( ( 1) 2 a + n 1 ) d (2d) n 1. 2 Note that the Smarandache bisymmetric determinant natural sequence is a particular case of the Smarandache bisymmetric arithmetic determinant sequence when a = 1 and d = 1. Open Problem: To find a formula for the sum of the first n terms of the Smarandache bisymmetric arithmetic determinant sequence. References [1] Amarnath Murthy, Smarandache Determinant Sequences, Smarandache Notions Journal, 12(2001), [2] A. A. K. Majumdar, On Some Smarandache Determinant Sequences, Scientia Magna, 4(2008), No.2,

92 Scientia Magna Vol. 4 (2008), No. 3, On the k-power complements function of n! Xiaoyan Li Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any fixed natural number k 2 and any positive integer n, we call b k (n) as a k-power complements function of n, if b k (n) denotes the smallest positive integer such that n b k (n) is a perfect k-power. In this paper, we use the elementary method to study the asymptotic properties of b k (n!), and give an interesting asymptotic formula for ln b k (n!). Keywords k-power complements function, Standard factorization, Asymptotic formula. 1. Introduction and results For any fixed natural number k 2 and any positive integer n, we call b k (n) as a k-power complements function of n, if b k (n) denotes the smallest positive integer such that n b k (n) is a perfect k-power. Especially, we call b 2 (n), b 3 (n), b 4 (n) as the square complements function, cubic complements function and quartic complements function respectively. In reference [1], Professor F.Smarandache asked us to study the properties of the k-power complements function. About this problem, there are many authors had studied it, and obtained a series results. For example, in reference [2], Wenpeng Zhang calculated the value of the series + n=1 1 (n b k (n)) s, where s is a complex number with Re(α) 1, k = 2, 3, 4 and prove that + n=1 1 (n b 2 (n)) s = ζ2 (2s) ζ(4s), where ζ(s) is the Riemann-zeta function. Ruiqin Fu [3] discussed the asymptotic properties of ln b 2 (n!), and proved that ( ( )) A ln 3 5 n ln b 2 (n!) = n ln 2 + O n exp, (ln ln n) 1 5 where n 2 is an positive integer, A > 0 is a constant. But for the properties of the k-power complements function of n, we still know very little at present. In this paper, we use the elementary method to study the asymptotic properties of b k (n!), and give an interesting asymptotic formula for ln b k (n!). That is, we shall prove the following general conclusion:

93 Vol.4 On the k-power complements function of n! 87 Theorem. For any fixed natural number k 2 and any positive integer n, we have the asymptotic formula ( ) ( ( )) 1 A ln 3 5 n ln b k (n!) = n k + O n exp, d(kd + 1) (ln ln n) 1 5 d=1 where A > 0 is a constant, and exp(y) = e y. 2. Two simple lemmas To complete the proof of the theorem, we need the following two simple lemmas: Lemma 1. Let n! = p α1 1 pα2 2 pαs s denotes the factorization of n! into prime powers. Then we have the calculate formula b k (n!) = b k (p α1 1 )b k(p α2 2 ) b k(p αs s ) = p ord(p1) 1 p ord(p2) 2 p ord(ps) s where the ord function is defined as: ord(p i ) = k 1, ifα i = km + 1, k 2, ifα i = km + 2,. 0, ifα i = k(m + 1). where m = 0, 1, 2,. Proof. See reference [3]. Lemma 2. For any real number x 2, we have the asymptotic formula θ(x) = ( ( )) A ln 3 5 x ln p = x + O x exp. (ln ln x) 1 5 p x where exp(y) = e y, and A > 0 is a constant. Proof. See reference [4] and [5]. 3. Proof of the theorem In this section, we shall complete the proof of Theorem. First from Lemma 1 we have ln b k (n!) = ln (b k (p α1 1 )b k(p α2 2 ) b k(p αs s )) = ord(p) ln p p n k ord(p) = (k 1) ln p + (k 2) ln p + + ln p n 2 <p n n 3 <p n n 2 k <p n k 1 + (k 1) ln p + + ln p + + O(1). n k+2 <p n k+1 n 2k <p n 2k 1

94 88 Xiaoyan Li No.3 Let n be a positive large enough, if a prime factor p of n! in the interval ( n, n], then the power 2 of p is 1 in the standard factorization of n!. Similarly, if a prime factor p of n! in the interval ( n 3, n ], then the power of p is 2 in the standard factorization of n!; If a prime factor p of n! in 2 the interval ( n 4, n ], then the power of p is 3 in the standard factorization of n!,. 3 Then from the above formula and Lemma 2 we have ( ( n )) ( ( n ) ( n )) ( ln b k (n!) = (k 1) θ(n) θ + (k 2) θ θ ( ( ) ( )) ( ( ) n n n + (k 1) θ θ + + θ θ k + 1 k + 2 2k 1 ( ( = n k )) + n ( k k k + 1 ( ( )) A ln 3 5 n + + O n exp (ln ln n) 1 5 ( M ) ( ( )) = lim n k M M kd A ln 3 5 n + O n exp d (ln ln n) 1 5 d=0 d=1 ( ) ( ( )) M = lim n 1 A ln 3 5 n k + O n exp M d(kd + 1) (ln ln n) 1 5 d=1 ( ) ( ( )) 1 A ln 3 5 n = n k + O n exp, d(kd + 1) (ln ln n) 1 5 d=1 ( ) n ( n ) ) θ θ k 1 k ( n ) ) + + O(1) 2k )) ( 1 k k k where A > 0 is a constant, and exp(y) = e y. This completes the proof of Theorem. k-power residues numbers. Absolutely, this theorem can be used to study References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] Zhang Wenpeng, Research on Smarandache Problems in Number Theory, Hexis, 2004, [3] Felice Russo, An introduction to the Smarandache square complements, Smarandache Notions Journal, 13(2002), [4] Chengdong Pan and Chengbiao Pan, Foundation of analytic number theory, Beijing: Science Press, [5] J.B.Rosser and L.Schoenfeld, Approximates formulas for some functions of prime numbers, Illions J.Math, 6(1962), [6] Zhang Tianping, On the cubic residues numbers and k-power complement numbers, Smarandache Notions Journal, 14(2004), [7] Tom M.Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.

95 Scientia Magna Vol. 4 (2008), No. 3, An alternative approach to the LP problem with equality constraints A.A.K. Majumdar APU, 1-1 Jumonjibaru, Beppu-shi , Oita-ken, Japan majumdar@apu.ac.jp aakmajumdar@gmail.com Abstract This paper gives an alternative approach to the solution of the LP problem with equality constraints. In this scheme, the constraint system of linear equations is first solved by elementary row operations (if such a solution exists). In the second stage, the solution of the system of linear equations is directly plugged in the objective function and then it is minimized. This scheme thus avoids the use of the simplex method. Keywords Linear programming, equality constraint, elementary row operations. 1. Introduction In a recent paper, Ru, Shen and Xue [1] considered the problem of finding an initial basic feasible solution (bfs) of the LP problem of the form min z = C t X s.t. AX = B, X 0, (1) where C = c 1 c 2. c n a 11., A =. a 21 a i1 a 12 a 1n a 22 a 2n a i2 a in, B = b 1 b 2. b m, X = x 1 x 2. x n. a m1 a m2 a mn In (1), without loss of generality, we may assume that B 0; for otherwise, if some b i < 0 (1 i m), then multiplying the i-th constraint throughout by (-1), we get one with ( b i ) > 0. To find an initial bfs of the LP problem (1), the procedure suggested by Ru et al. [1] is as follows : Step 1 : Starting with the augmented matrix (A B), reduce it to a form with an identity matrix of desired order, by elementary row operations only. Note that, the elementary row operations applicable in this case are (a) multiplying any row of (A B) by a positive constant,

96 90 A.A.K. Majumdar No. 3 (b) multiplying any row of (A B) by a constant and adding it to a second row. Throughout this paper, the first row operation would be denoted by R i kr i (which means that the i-th row R i is multiplied by the constant k, where k > 0), and the second row operation would be denoted by R i R i + kr j (which means that the j-th row R j is multiplied by the constant k and the resulting row is added, term-by-term, to the i-th row R i to get the new i-th row). Note that the latter row operation is allowed only if b i + kb j 0. If rank(a) = rank(a B) = l( m), then using the above elementary row operations, it is possible to find a unit matrix of order l, which may be exploited to find an initial bfs of the LP problem (1). It may be mentioned here that, if l < m, then only l of the m system of linear equations AX = B are independent, and the remaining l m equations can be ignored. It may also be mentioned here that, if rank(a) rank(a B), then the system of linear equations AX = B is inconsistent, and as such, the LP problem (1) is infeasible. Step 2 : With the initial bfs found in Step 1 above, form the initial simplex tableau in the usual manner. Step 3 : Proceed in the usual way of the simplex method to find an optimal solution. The method outlined above is nevertheless not a new one. In fact, the same procedure has been followed by Papadimitriou and Steiglitz [2]. In the next section, we propose an alternative approach. 2. An alternative approach In this section, we suggest an alternative approach to the solution of the LP problem (1). The procedure is quite simple, and can be done in the following two steps. Step 1 : Reduce the system of linear equations AX = B to a form containing the unit matrix of requisite order, using elementary row operations only. Step 2 : With the solution found in Step 1, consider the problem of minimizing z = C t X. Thus, the method works with the solution of the system of linear equations AX = B and the objective function only, avoiding the simplex method completely. We illustrate our method with the help of the following examples, due to Ru et al. [1]. Example 2.1. Consider the following LP problem : min z = 4x 1 + 3x 2 s.t. 1 2 x 1 +x x x 4 = x x 3 = 3 3x 1 6x 2 +4x 4 = 0 x 1, x 2, x 3, x 4 0. To solve the above LP problem, we first solve the constraint system of linear equations, using elementary row operations only. We thus start with the augmented matrix (A B) and

97 Vol. 4 An alternative approach to the LP problem with equality constraints 91 proceed as follows, using the indicated row operations: (A B) = R 3 R 3 + 6R R 1 R 1 + R 2 R 2 R R R R 3 R 1 R R 2 R 3 R R The final tableau gives the following solution : x 1 = 2, x 3 = 0, x 2 = x 4 = t; (t 0 is any real number), 3 and the objective function is z = 4x 1 + 3x 2 = 8 + 3( t) = t. Since the objective is to minimize z, it is clear that z is minimized when t = 0, and the optimal solution of the given LP problem is thus x 1 = 2, x 2 = 1, x 3 = 0, x 4 = 0; min z = 11. Example 2.2. To solve the LP problem min z = 3x 1 + x 2 + x 3 s.t. x 1 2x 2 +x 3 +x 4 = 11 4x 1 +x 2 +2x 3 x 5 = 3 2x 1 +x 3 = 1 x 1, x 2, x 3, x 4, x 5 0, we first solve the constraint system of linear equations by elementary row operations. This is done below (A B) = R 1 R 1 + 2R R 1 R 1 + 5R The final tableau is equivalent to the following system of linear equations with the solution 3x 1 +x 4 2x 5 = 12 x 2 x 5 = 2 2x 1 +x 3 = 1 x 1 = t 1, x 5 = t 2, x 2 = 1+t 2, x 3 = 1+2t 1, x 4 = 12 3t 1 +2t 2 ; (t 1 0, t 2 0 are any real numbers),

98 92 A.A.K. Majumdar No. 3 and the objective function is z = 2 t 1 + t 2. Since the objective is to minimize z, it is evident that t 2 should be as small as possible and t 1 should be as large as possible. Now, since x 4 = 12 3t 1 + 2t 2 0, we see that, we must have t 2 = 0, t 1 = 4. Hence, the desired optimal solution is x 1 = 4, x 2 = 1, x 3 = 9, x 4 = 0, x 5 = 0; min z = 3. Example 2.3. To solve the LP problem min z = 4x 1 + 3x 3 s.t. 1 2 x 1 +x x x 4 = x x 3 = 3 3x 1 6x 2 +x 4 = 0 x 1, x 2, x 3, x 4 0, we start with the augmented matrix (A B) and proceed as follows, using the indicated row operations : (A B) = R R 3 + 6R R 3 R 3 4R 3 3 R 1 R R R R The final tableau is equivalent to the following system of linear equations with the solution x x x 4 = 1 x x 3 = 2 x 3 = t 1, x 1 = t 1, x 4 = t 2, x 2 = t t 2; (t 1 0, t 2 0 are any real numbers), and the objective function is z = 8 + t 1. Since the objective is to minimize z, we must have t 1 = 0. Hence, the desired solution is x 1 = 2, x 2 = t, x 3 = 0, x 4 = t(t 0 is any real number); min z = 8. The above solution shows that the given LP problem has infinite number of solutions.

99 Vol. 4 An alternative approach to the LP problem with equality constraints Some observations and remarks In the previous section, we have illustrated our method with the help of three simple examples. However, in large-scale problems, the success as well as the computational efficiency of the method remains to be checked out. Clearly, the method is useful when the correct basis has been found. This point is illustrated with the help of the following example, due to Papadimitriou and Steiglitz [2]. Example 3.1. Consider the LP problem below : min z = x 1 + x 2 + x 3 + x 4 + x 5 s.t. 3x 1 +2x 2 +x 3 = 1 5x 1 +x 2 +x 3 +x 4 = 3 2x 1 +5x 2 +x 3 +x 5 = 4 x 1, x 2, x 3, x 4, x 5 0. We first solve the constraint system of linear equations by elementary row operations, using the indicated row operations R R (A B) = R 2 R 2 R R 3 R 3 5R The final tableau is equivalent to the following system of linear equations with the solution 3 2 x 1 +x x 3 = x x 3 +x 4 = x x 3 +x 5 = 3 2 x 1 = t 1, x 3 = t 2 (t 1 0, t 2 0 are any real numbers), x 2 = t t 2, x 4 = t t 2, x 5 = t t 2; and the objective function is z = (t 1 + t 2 ). Since the objective is to minimize z, it is evident that we must have t 1 = 0, t 2 = 0, giving the desired optimal solution x 1 = 0, x 2 = 1 2, x 3 = 0, x 4 = 5 2, x 5 = 3 2 ; min z = 9 2. On the other hand, the solution below (A B) = R 2 R 2 R 1 R 3 R 3 R

100 94 A.A.K. Majumdar No. 3 leads to the system of equations whose solution is 3x 1 +2x 2 +x 3 = 1 2x 1 x 2 +x 4 = 2 x 1 +3x 2 +x 5 = 3 x 1 = t 1, x 2 = t 2 (t 1 0, t 2 0 are any real numbers), x 3 = 1 3t 1 2t 2, x 4 = 2 2t 1 + t 2, x 5 = 3 + t 1 3t 2 ; and the objective function is z = 3[2 (t 1 + t 2 )]. Thus, the problem of finding the solution of the given LP problem reduces to the problem of solving the LP problem max t 1 + t 2 s.t. 3t 1 +2t 2 1 2t 1 t 2 2 t 1 +3t 2 3 t 1, t 2 0. The optimal solution of the above LP problem is t 1 = 0, t 2 = 1 2. Hence, the optimal solution of the original LP problem is x 1 = 0, x 2 = 1 2, x 3 = 0, x 4 = 5 2, x 5 = 3 2 ; min z = 9 2. The above example shows that, in the alternative approach proposed in this paper, as well as the method suggested by Ru et al. [1], the choice of the unit matrix (of desired order) during the solution of the system of linear equations AX = B (by elementary row operations) may save the number of iterations considerably. However, no particular criterion is specified in this respect. It thus still remains to check how far these methods can compete with the existing method. References [1] Shaofeng Ru, Maoxing Shen and Xifeng Xue, A Concise Way of Determination for LP Initial Feasible Basis of Simplex Method, Scientia Magna, 4(2008), No.1, [2] Christos H. Papadimitriou and Kenneth Steiglitz, Combinatorial Optimization : Algorithms and Complexity, Dover Publications, Inc., 1998.

101 Scientia Magna Vol. 4 (2008), No. 3, On some pseudo Smarandache function related triangles A.A.K. Majumdar APU, 1-1 Jumonjibaru, Beppu-shi , Oita-ken, Japan majumdar@apu.ac.jp aakmajumdar@gmail.com Abstract Two triangles T (a, b, c) and T (a, b, c ) are said to be pseudo Smarandache related if Z(a) = Z(a ), Z(b) = Z(b ), Z(c) = Z(c ), where Z(.) is the pseudo Smarandache function, and T (a, b, c) denotes the triangle with sides a, b and c. This paper proves the existence of an infinite family of pairs of dissimilar Pythagorean triangles that are pseudo Smarandache related. Keywords Smarandache function, pseudo Smarandache function, Smarandache related triangles, pseudo Smarandache related triangles, Pythagorean triangles. 1. Introduction and result The Smarandache function, denoted by S(n), is defined as follows. Definition 1.1. For any integer n 1, ( Z + being the set of all positive integers), S(n) = min{m : m Z +, n m!}. The following definition is due to Sastry [6]. Definition 1.2. Two triangles T (a, b, c) (with sides of length a, b and c) and T (a, b, c ) (with sides of length a, b and c ), are said to be Smarandache related if S(a) = S(a ), S(b) = S(b ), S(c) = S(c ). Definition 1.3. A triangle T (a, b, c) is said to be Pythagorean if and only if one of its angles is 90. Thus, a triangle T (a, b, c) (with sides of length a, b and c) is Pythagorean if and only if a 2 + b 2 = c 2. Sastry[6] raised the following question : Are there two distinct dissimilar Pythagorean triangles that are Smarandache related? Recall that two triangles T (a, b, c) and T (a, b, c ) are similar if and only if the corresponding three sides are proportional, that is, if and only if a a = b b = c c Otherwise, the two triangles are dissimilar.

102 96 A.A.K. Majumdar No. 3 The following result, due to Ashbacher [1], answers the question in the affirmative. Theorem 1.1. There are an infinite family of pairs of dissimilar Pythagorean triangles that are Smarandache related. The proof of Theorem 1.1 is rather simple : For any prime p 17, the two families of dissimilar Pythagorean triangles T (3p, 4p, 5p) and T (5p, 12p, 13p) are Smarandache related, since S(3p) = S(4p) = S(5p) = S(12p) = S(13p) = p. Ashbacher [1] introduced the concept of pseudo Smarandache related triangles, defined as follows. Definition 1.4. Two triangles T (a, b, c) and T (a, b, c ) are said to be pseudo Smarandache related if Z(a) = Z(a ), Z(b) = Z(b ), Z(c) = Z(c ). In Definition 1.4 above, Z(.) denotes the pseudo Smarandache function. Recall that the pseudo Smarandache function is defined as follows. Definition 1.5. For any integer n 1, Z(n) is the smallest positive integer m such that m(m + 1) m is divisible by n. Thus, 2 Z(n) = min{m : m Z +, n m(m + 1) }; n 1. 2 Ashbacher [1] used a computer program to search for dissimilar pairs of Pythagorean triangles that are pseudo Smarandache related. He reports some of them, and conjectures that there are an infinite number of pairs of dissimilar Pythagorean triangles that are pseudo Smarandache related. In this paper, we prove the conjecture of Ashbacher [1] in the affirmative. This is done in Theorem 3.1 in Section 3. We proceed on the same line as that followed by Ashbacher. However, in the case of the pseudo Smarandache related triangles, the proof is a little bit more complicated. The intermediate results, needed for the proof of Theorem 3.1, are given in the next Section Some preliminary results The following result, giving the explicit expressions of Z(3p), Z(4p) and Z(5p), are given in Majumdar [5]. Lemma 2.1. If p 5 is a prime, then p 1, if 3 (p 1), Z(3p) = p, if 3 (p + 1). Lemma 2.2. If p 5 is a prime, then p 1, p, Z(4p) = 3p 1, 3p, if 8 (p 1), if 8 (p + 1), if 8 (3p 1), if 8 (3p + 1).

103 Vol. 4 On some pseudo Smarandache function related triangles 97 Lemma 2.3. If p 7 is a prime, then p 1, p, Z(5p) = 2p 1, 2p, if 10 (p 1), if 10 (p + 1), if 5 (2p 1), if 5 (2p + 1). The explicit expressions of Z(12p) and Z(13p) are given in the following two lemmas. Lemma 2.4. If p 13 is a prime, then p 1, if 24 (p 1) p, if 24 (p + 1), 3p 1, if 8 (3p 1), Z(12p) = 3p, if 8 (3p + 1), 7p 1, if 24 (7p 1), 7p, if 24 (7p + 1). Proof. By definition, m(m + 1) m(m + 1) Z(12p) = min{m : 12p } = min{m : p }. (1) 2 24 If p m(m + 1), then p must divide either m or m + 1, but not both, and then 24 must divide either m + 1 or m respectively. In the particular case when 24 divides p 1 or p + 1, the minimum m in (1) may be taken as p 1 or p respectively. We now consider the following eight cases that may arise : Case 1 : p is of the form p = 24a + 1 for some integer a 1. In this case, 24 (p 1). Therefore, Z(12p) = p 1. Case 2 : p is of the form p = 24a + 23 for some integer a 1. Here, 24 (p + 1), and so, Z(12p) = p. Case 3 : p is of the form p = 24a + 5 for some integer a 1, so that 8 (3p + 1). In this case, the minimum m in (1) may be taken as 3p. That is, Z(12p) = 3p. Case 4 : p is of the form p = 24a + 19 for some integer a 1. Here, 8 (3p 1), and hence, Z(12p) = 3p 1. Case 5 : p is of the form p = 24a + 7 for some integer a 1. In this case, 24 (7p 1), and hence, Z(12p) = 7p. Case 6 : p is of the form p = 24a + 17 for some integer a 1. Here, 24 (7p + 1), and hence, Z(12p) = 7p + 1. Case 7 : p is of the form p = 24a + 11 for some integer a 1. In this case, 8 (3p 1), and hence, Z(12p) = 3p - 1. Case 8 : p is of the form p = 24a + 13 for some integer a 1. Here, 8 (3p + 1), and hence, Z(12p) = 3p.

104 98 A.A.K. Majumdar No. 3 Lemma 2.5. For any prime p 17, p 1, p, 2p 1, 2p, 3p 1, 3p, Z(13p) = 4p 1, 4p, 5p 1, 5p, 6p 1, 6p, if 13 (p 1), if 13 (p + 1), if 13 (2p 1), if 13 (2p + 1), if 13 (3p 1), if 13 (3p + 1), if 13 (4p 1), if 13 (4p + 1), if 13 (5p 1), if 13 (5p + 1), if 13 (6p 1), if 13 (6p + 1). Proof. By definition, m(m + 1) m(m + 1) Z(13p) = min{m : 13p } = min{m : p }. (2) 2 26 We have to consider the twelve possible cases that may arise : Case 1 : p is of the form p = 13a + 1 for some integer a 1. In this case, 13 (p 1), and so, Z(13p) = p 1. Case 2 : p is of the form p = 13a + 12 for some integer a 1. Here, 13 (p + 1), and hence, Z(13p) = p. Case 3 : p is of the form p = 13a + 2 for some integer a 1. In this case, 13 (6p + 1), and hence, Z(13p) = 6p. Case 4 : p is of the form p = 13a + 11 for some integer a 1. Here, 13 (6p 1), and hence, Z(13p) = 6p 1. Case 5 : p is of the form p = 13a + 3 for some integer a 1. In this case, 13 (4p + 1), and hence, Z(13p) = 4p. Case 6 : p is of the form p = 13a + 10 for some integer a 1. Here, 13 (4p 1), and hence, Z(13p) = 4p 1. Case 7 : p is of the form p = 13a + 4 for some integer a 1. In this case, 13 (3p + 1), and hence, Z(13p) = 3p. Case 8 : p is of the form p = 13a + 9 for some integer a 1. Here, 13 (3p 1), and hence, Z(13p) = 3p 1. Case 9 : p is of the form p = 13a + 5 for some integer a 1. In this case, 13 (5p + 1), and hence, Z(13p) = 5p. Case 10 : p is of the form p = 13a + 8 for some integer a 1. Here, 13 (5p 1), and hence, Z(13p) = 5p 1. Case 11 : p is of the form p = 13a + 6 for some integer a 1. In this case, 13 (2p + 1), and hence, Z(13p) = 2p.

105 Vol. 4 On some pseudo Smarandache function related triangles 99 Case 12 : p is of the form p = 13a + 7 for some integer a 1. Here, 13 (2p 1), and hence, Z(13p) = 2p Main result We are now state and prove the main result of this paper in the following theorem. Theorem 3.1. There are an infinite number of pairs of dissimilar Pythagorean triangles that are pseudo Smarandache related. Proof. We consider the pair of dissimilar Pythagorean triangles where p is a prime of the form By Lemma Lemma 2.5, T (3p, 4p, 5p) and T (5p, 12p, 13p), (3) p = ( )n + 1 = 1560n + 1, n Z +. (4) Z(3p) = Z(4p) = Z(5p) = Z(12p) = Z(13p) = p 1, so that the triangles T (3p, 4p, 5p) and T (5p, 12p, 13p) are pseudo Smarandache related. Now, since there are an infinite number of primes of the form (4) (by Dirichlet s Theorem, see, for example, Hardy and Wright [3], Theorem 15, pp. 13), we get the desired infinite number of pairs of dissimilar Pythagorean triangles that are pseudo Smarandache related. It may be mentioned here that, the family of pairs of triangles (3), where p is a prime of the form p = 1560n 1, n Z +, (5) also forms (dissimilar) pseudo Smarandache related Pythagorean triangles. 4. Some remarks The pseudo Smarandache function Z(n) is clearly not bijective. However, we can define the inverse Z 1 (m) as follows : with Z 1 (m) = {n Z + : Z(n) = m} for any integer m 3, (6) Z 1 (1) = 1, Z 1 (2) = 3. (7)

106 100 A.A.K. Majumdar No. 3 As has been pointed out by Majumdar [5], for any m Z +, the set Z 1 (m) is non-empty m(m + 1) and bounded with as its largest element. Clearly, n Z 1 (m) if and only if the 2 following two conditions are satisfied : m(m + 1) (1) n divides, 2 l(l + 1) (2) n does not divide for any l with 1 l m 1. 2 We can look at (6) from a different point of view : On the set Z +, we define the relation R as follows : For any n 1, n 2 Z +, n 1 R n 2 if and only if Z(n 1 ) = Z(n 2 ). (8) It is then straightforward to verify that R is an equivalence relation on Z +. It is wellknown that an equivalence relation induces a partition (on the set Z + ) (see, for example, Gioia [2], Theorem 11.2, pp. 32). The sets Z 1 (m), m Z +, are, in fact, the equivalence classes induced by the equivalence relation R on Z +, and possesses the following two properties : (1) Z 1 (m) = Z + m=1 (2) Z 1 (m 1 ) Z 1 (m 2 ) =, if m 1 m 2. Thus, for any n Z +, there is one and only one m Z + such that n Z 1 (m). A different way of relating two triangles has been considered by Ashbacher [1], which is given in the following definition. Definition 4.1. Given two triangles, T (a, b, c) and T (a, b, c ), where they are said to be pseudo Smarandache related if a + b + c = 180 = a + b + c, (9) Z(a) = Z(a ), Z(b) = Z(b ), Z(c) = Z(c ). The difference in Definition 1.4 and Definition 4.1 is that, in the former, the sides of the pair of triangles are pseudo Smarandache related, while their angles, measured in degrees, are pseudo Smarandache related in the latter case. Note that the condition (9) merely states the fact the sum of the three angles of a triangle is 180 degrees. Using a computer program, Ashbacher searched for pseudo Smarandache related pairs of triangles (in the sense of Definition 4.1). He reports three such pairs. However, in this case, the equivalence classes Z 1 (m), m Z +, might be of some help. The condition (9) can be dealt with by considering the restricted sets Z 1 (m π) : Z 1 (m π) = {n Z + : Z(n) = m, 1 n 178}. (10) Table 4.1 gives such restricted sets related to our problem. Clearly, two pseudo Smarandache related triangles T (a, b, c) and T (a, b, c ) must satisfy the following condition : a, a Z 1 (m 1 π); b, b Z 1 (m 2 π), c, c Z 1 (m 3 π) for some m 1, m 2, m 3 Z +.

107 Vol. 4 On some pseudo Smarandache function related triangles 101 Thus, for example, choosing a, a {2, 6} = Z 1 (3 π); b, b {44, 48, 88, 132, 176} = Z 1 (32 π), we can construct the pseudo Smarandache related triangles T (2, 48, 130) and T (6, 44, 130). Again, choosing a, a, b, b {25, 50, 75, 100, 150} = Z 1 (24 π), we can form the pair of pseudo Smarandache related triangles T (25, 150, 5) and T (75, 100, 5) with the characteristic that Z(a) = Z(a ) = 24 = Z(b) = Z(b ) (and Z(c) = Z(c ) = 4). Choosing a, a, b, b {8, 20, 24, 30, 40, 60, 120} = Z 1 (15 π), we see that the equilateral triangle T (60, 60, 60) is pseudo Smarandache related to the triangle T (20, 40, 120)! Ashbacher [1] cites the triangles T (4, 16, 160) and T (14, 62, 104) as an example of a pseudo Smarandache related pair where all the six angles are different. e have found three more, given below : (1) T (8, 12, 160) and T (40, 36, 104), (with Z(8) = Z(40) = 15, Z(12) = Z(36) = 8, Z(160) = Z(104) = 64), (2) T (16, 20, 144) and T (124, 24, 32), (with Z(16) = Z(124) = 31, Z(20) = Z(24) = 15, Z(144) = Z(32) = 63), (3) T (37, 50, 93) and T (74, 75, 31), (with Z(37) = Z(74) = 36, Z(50) = Z(75) = 24, Z(93) = Z(31) = 30). Table 4.1. Values of Z 1 (m π) = {n Z + : Z(n) = m, 1 n 180}

108 102 A.A.K. Majumdar No. 3 m Z 1 (m π) m Z 1 (m π) m Z 1 (m π) 1 {1} 35 {90} 80 {81, 108, 162, 180} 2 {3} 36 {37, 74, 111} 82 {83} 3 {2, 6} 39 {52, 130, 156} 83 {166} 4 {5, 10} 40 {41, 82, 164} 84 {170} 5 {15} 41 {123} 87 {116, 174} 6 {7, 21} 42 {43, 129} 88 {89, 178} 7 {4, 14, 28} 43 {86} 95 {152} 8 {9, 12, 18, 36} 44 {99, 110, 165} 96 {97} 9 {45} 45 {115} 100 {101} 10 {11, 55} 46 {47} 102 {103} 11 {22, 33, 66} 47 {94, 141} 106 {107} 12 {13, 26, 39, 78} 48 {49, 56, 84, 98, 147, 168} 108 {109} 13 {91} 49 {175} 111 {148} 14 {35, 105} 51 {102} 112 {113} 15 {8, 20, 24, 30, 40, 60, 120} 52 {53, 106} 120 {121} 16 {17, 34, 68, 136} 53 {159} 124 {125} 17 {51, 153} 54 {135} 126 {127} 18 {19, 57, 171} 55 {140, 154} 127 {64} 19 {38, 95} 56 {76, 114, 133} 128 {172} 20 {42, 70} 58 {59} 130 {131} 21 {77} 59 {118, 177} 136 {137} 22 {23} 60 {61, 122} 138 {139} 23 {46, 69, 92, 138} 63 {32, 72, 96, 112, 144} 148 {149} 24 {25, 50, 75, 100, 150} 64 {80, 104, 160} 150 {151} 25 {65} 65 {143} 156 {157} 26 {27, 117} 66 {67} 162 {163} 27 {54, 63, 126} 67 {134} 166 {167} 28 {29, 58} 69 {161} 168 {169} 29 {87, 145} 70 {71} 172 {173} 30 {31, 93, 155} 71 {142} 178 {179} 31 {16, 62, 124} 72 {73, 146} 255 {128} 32 {44, 48, 88, 132, 176} 78 {79} 34 {85, 119} 79 {158} Ashbacher [1] reports that, an exhaustive computer search for pairs of dissimilar search for

109 Vol. 4 On some pseudo Smarandache function related triangles 103 all pseudo Smarandache related triangles T (a, b, c) and T (a, b, c ) (a+b+c = 180 = a +b +c ) with values of a in the range 1 a 178, revealed that a cannot take the following values : 1, 15, 23, 35, 41, 45, 51, 59, 65, 67, 71, 73, 77, 79, 82, 83, 86, 87, 89, (11) 90, 91, 97, 101, 102, 105, 107, 109, 113, 115, 116, 118, 121, 123, 125, 126, 127, 131, 134, 135, 137, 139, 141, 142, 143, 148, 149, 151, 152, 153, 157, 158, 159, 161, 163, 164, 166, 167, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178. Can the table of the sets Z 1 (m π), Table 4.1, be utilized in explaining this observation? Of course, if a is too large, then there is a possibility that no two dissimilar triangles exist: Very large value of a very often forces the two triangles to be similar. For example, if a = 174 Z 1 (87 π), then the three possible pairs of values of (b, c) are (1, 5), (2, 4) and (3, 3). Note that the two sets Z 1 (1 π) = {1} and Z 1 (2 π) = {3} are singleton. Thus, if a = 174, then the two triangles T (a, b, c) and T (a, b, c ) must be similar. On the other hand, if a = 116 Z 1 (87 π), then we cannot find b, c with b + c = 64. Now, we consider the case when a = 164. This case is summarized in the tabular form below : a=164 a =164 a =82 a =41 Remark (1,15) Similar Triangle Not Possible Not Possible Both belong to singleton sets (b,c) (2,14) Similar Triangle Not Possible Not Possible b + c 34 (3,13) Similar Triangle Not Possible Not Possible 3 belongs to a singleton set (4,12) Similar Triangle Not Possible Not Possible b + c 64 (5,11) Similar Triangle Not Possible Not Possible b + c 65 (6,10) Similar Triangle Not Possible Not Possible b + c 16 (7,9) Similar Triangle Not Possible Not Possible b + c 57 (8,8) Similar Triangle Not Possible Not Possible However, if a = 165, we can get two dissimilar pseudo Smarandache related triangles, namely, T (165, 7, 8) and T (81, 21, 60). A closer look at the values listed in (11) and those in Table 4.1 reveals the following facts: (1) The numbers not appearing in any 180 degrees triplets are all belong to singleton sets, with the exception of 3 (though we conjecture that 3 cannot appear in any triplet), 47, 64, and 103. For the last three cases, we can form the following examples : (a) T (47, 76, 57) and T (47, 114, 19) (with Z(76) = 56 = Z(114), Z(57) = 18 = Z(19)), (b) T (64, 12, 104) and T (64, 36, 80) (with Z(12) = 8 = Z(36), Z(104) = 64 = Z(80)), (c) T (103, 11, 66) and T (103, 55, 22) (with Z(11) = 10 = Z(55), Z(66) = 11 = Z(22)). (2) In most of the cases, if a value does not appear in the triplet, all other values of the corresponding Z 1 (m π) also do not appear in other triplet, with the exception of 145 Z 1 (29 π) (87 Z 1 (29 π) does not appear in any triplet), and 146 Z 1 (72 π)(73 Z 1 (72 π) does not appear in any triplet). In this connection, we may mention the following pairs of triplets :

110 104 A.A.K. Majumdar No. 3 (a) T (145, 14, 21) and T (145, 28, 7) (with Z(14) = 7 = Z(28), Z(21) = 6 = Z(7)), (b) T (146, 4, 30) and T (146, 14, 20) (with Z(4) = 7 = Z(14), Z(30) = 15 = Z(20)). (c) Both 54, 63 Z 1 (72 π) can appear in one or the other triplet, but 126 Z 1 (72 π) cannot appear in any triplet. It is also an interesting problem to look for 60 degrees and 120 degrees pseudo Smarandache related pairs of triangles, in the sense of Definition 4.1, where a 60 (120) degrees triangle is one whose one angle is 60 (120) degrees. We got the following two pairs of pseudo Smarandache related 60 degrees triangles : (1) T (8, 60, 112) and T (24, 60, 96), (with Z(8) = Z(24) = 15, Z(112) = Z(96) = 63), (2) T (32, 60, 88) and T (72, 60, 48), (with Z(32) = Z(72) = 63, Z(88) = Z(48) = 32), while our search for a pair of pseudo Smarandache related 120 degrees triangles went in vain. We conclude the paper with the following open problems and conjectures. The first two have already been posed by Ashbacher [1]. Problem 1. Are there an infinite number of pairs of dissimilar 60 degrees triangles that are pseudo Smarandache related, in the sense of Definition 1.4? Problem 2. Is there an infinite family of pairs of dissimilar 120 degrees triangles that are pseudo Smarandache related, in the sense of Definition 1.4? Ashbacher [1] reports that a limited search on a computer showed only four pairs of dissimilar 60 degrees psedo Smarandache related triangles, while the number is only one in the case of 120 degrees triangles. In the formulation of Definition 4.1, the number of pairs of dissimilar Smarandache related or pseudo Smarandache related triangles is obviously finite. In fact, since the number of integer solutions to a + b + c = 180 a 1, b 1, c 1 is C(179, 2) (see, for example, Johnsonbaugh [4], Theorem 4.5, pp. 238), the number of such triangles cannot exceed C(179, 2). But this is a very crude estimate. The next problems of interest are Problem 3. Is it possible to find a tight or better upper limit to the number of pairs of dissimilar triangles that are pseudo Smarandache related, in the sense of Definition 4.1? Problem 4. Is it possible to find a good upper limit to the number of pairs of dissimilar 60 degrees triangles that are pseudo Smarandache related, in the sense of Definition 4.1? Problem 5. Is it possible to find another pair of 60 degrees pseudo Smarandache related triangles, in the sense of Definition 4.1, where all the six angles are acute? By a limited search, we found only two pairs of dissimilar 60 degrees pseudo Smarandache related triangles, the ones already mentioned above. Conjecture 1. There is no pair of dissimilar 120 degrees triangles that are Pseudo Smarandache related, in the sense of Definition 4.1. Conjecture 2. There is no pair of dissimilar 90 degrees triangles that are Pseudo Smarandache related, in the sense of Definition 4.1.

111 Vol. 4 On some pseudo Smarandache function related triangles 105 References [1] Ashbacher, Charles, Solutions to Some Sastry Problems on Smarandache Number Related Triangles, Smarandache Notions Journal, 11 (2000), [2] Gioia, A. A., The Theory of Numbers, An Introduction, Dover Publications, Inc., [3] Hardy, G.H. and Wright, E.M., An Introduction to the Theory of Numbers, Oxford Science Publications, Clarendon Press, Oxford, UK 5 th Edition, [4] Johnsonbaugh, R., Discrete Mathematics, Prentice-Hall, Inc. London, UK, 4 th Edition, [5] Majumdar, A. A. K., A Note on the Pseudo Smarandache Function, Scientia Magna, 2 (2006), No.3, [6] Sastry, K. R. S., Smarandache Number Related Triangles, Smarandache Notions Journal, 11(2000),

112 Scientia Magna Vol. 4 (2008), No. 3, One problem related to the Smarandache function Wei Qin Department of Mathematics, Northwest University, Xi an, Shaanxi, P.R.China Abstract For any positive integer n, the famous F.Smarandache function S(n) is defined as the smallest positive integer m such that n m!. That is, S(n) = min{m : n m!, n N}. The main purpose of this paper is using the elementary method to study the number of the solutions of the congruent equation 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) ( mod n), and give its all prime number solutions. Keywords F. Smarandache function, divisibility, primitive root. 1. Introduction and result For any positive integer n, the famous F. Smarandache function S(n) is defined as the smallest positive integer m such that n m!. That is, S(n) = min{m : n m!, n N}. For example, the first few values of S(n) are S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 4, S(5) = 5, S(6) = 3, S(7) = 7, S(8) = 4, S(9) = 6, S(10) = 5, S(11) = 11, S(12) = 4,. About the elementary properties of S(n), many authors had studied it, and obtained a series interesting results, see references [1], [2], [3] and [4]. For example, Xu Zhefeng [2] studied the value distribution problem of S(n), and proved the following conclusion: Let P (n) denotes the largest prime factor of n, then for any real number x > 1, we have the asymptotic formula n x (S(n) P (n)) 2 = 2ζ ( ) x 2 3 ln x where ζ(s) is the Riemann zeta-function. Lu Yaming [3] studied the solutions of an equation involving the F. Smarandache function S(n), and proved that for any positive integer k 2, the equation + O ( x 3 2 ln 2 x S(m 1 + m m k ) = S(m 1 ) + S(m 2 ) + + S(m k ) ), has infinite group positive integer solutions (m 1, m 2,, m k ). Rongji Chen [4] proved that for any fixed r N with r 3, the positive integer n is a solution of the equation S(n) r + S(n) r S(n) = n

113 Vol. 4 One problem related to the Smarandache function 107 if and only if n = p(p r 1 + p r ), where p is an odd prime such that p r 1 + p r (p 1)!. On the other hand, in reference [5], C. Dumitrescu and V. Seleacu asked us to study the solvability of the congruent equation 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) mod n. (1) About this problem, it seems that none had studied it yet, at least we have not seen any related papers before. In this paper, we use the elementary method and the properties of the primitive roots to study the solvability of the congruent equation (1), and obtain its all prime solutions. That is, we shall prove the following conclusion: Theorem. Let n be a prime, then n satisfy the congruent equation 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) mod n if and only if n = 2, 3 and 5. It is clear that our Theorem obtained all prime solutions of the congruent equation (1). About the general solutions of the congruent equation (1) is still an unsolved problem. 2. Proof of the theorem In this section, we shall complete the proof of our Theorem directly. We only discuss the prime solutions of (1). (1) For n = 2, since 2 1 S(1) + 1 = 2, so n = 2 is a prime solution of (1). (2) For n = 3, since 3 1 S(2) + 2 S(2) + 1 = 6, so n = 3 is a prime solution of (1). (3) For n = 5, since 5 1 S(4) + 2 S(4) + 3 S(4) + 4 S(4) + 1 = 355, so n = 5 is also a prime solution of (1). (4) For prime n = p 7, it is clear that p has at least a primitive root. Let g be a primitive root of p, that is to say, ( g i 1, p ) = 1 for all 1 i p 2, the congruences g p 1 1 mod p and g m(p 1) 1 mod p (2) hold for any positive integer m. Then from the properties of the primitive root mod p we know that g 0, g 1,, g p 2 is a reduced residue class. So we have the congruent equation 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) (3) g 0 S(p 1) + g 1 S(p 1) + + g (p 2) S(p 1) g(p 1) S(p 1) 1 g S(p 1) 1 mod p.

114 108 Wei Qin No. 3 It is clear that for any prime p 7, we have S(p 1) p 2, so ( g S(p 1) 1, p ) = 1. Therefore, from (2) and (3) we have 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) g(p 1) S(p 1) 1 g S(p 1) 1 So from this congruence we may immediately get 1 S(n 1) + 2 S(n 1) + + (n 1) S(n 1) mod p. 0 mod p. Thus, if prime p 7, then it is not a solution of (1). This completes the proof of Theorem. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, [2] Xu Zhefeng, The value distribution property of the Smarandache function, Acta Mathematica Sinica, Chinese Series, 49 (2006), No.5, [3] Lu Yaming, On the solutions of an equation involving the Smarandache function, Scientia Magna, 2(2006), No.1, [4] Rongji Chen, On the functional equation S(n) r +S(n) r 1 + +S(n) = n, Smaraandache Notions Journal, 11(2000), No.1-2-3, [5] C. Dumitrescu and V. Seleacu, The Smarandache Function, Erhus University Press, USA, [6] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi an, [7] Tom M.Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, [8] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, 1996.

115 Scientia Magna Vol. 4 (2008), No. 3, Notes on mssc-images of relatively compact metric spaces Nguyen Van Dung Department of Mathematics, Pedagogical University of Dongthap, Caolanh City, Dongthap Province, Vietnam nguyendungtc@yahoo.com, nvdung@pud.edu.vn Abstract We introduce the notion of σ-finite Ponomarev-system (f, M, X, {P n}) to give a consistent method to construct an mssc-mapping f with covering-properties onto a space X from some relatively compact metric space M. As applications, we systematically obtain characterizations on mssc-images of relatively compact metric spaces under certain coveringmappings, which sharpen results in [18, 26], and more. Keywords σ-finite Ponomarev-system, relatively compact metric, sn-network, so-network, cs-network, k-network, 2-sequence-covering, 1-sequence-covering, sequence-covering, compactcovering, mssc-mapping. 1. Introduction An investigation of relations between spaces with countable networks and images of separable metric spaces is one of interesting questions on generalized metric spaces. Related to characterizing a space X having a certain countable network P by an image of a separable metric space M under some covering-mapping f, many results have been obtained. In the past, E. Michael [18] proved the following results. Theorem 1.1. ([18], Proposition 10.2) The following are equivalent for a space X. (1) X is a cosmic space. (2) X is an image of a separable metric space. Theorem 1.2. ([18], Theorem 11.4) The following are equivalent for a space X. (1) X is an ℵ 0 -space. (2) X is a compact-covering image of a separable metric space. Also, he posed the following question, on page 999. Question 1.3. Find a better explanation for images of separable metric spaces? In [26], Y. Tanaka and Z. Li proved the following result. Theorem 1.4. ([26], Corollary 8 (1)) The following are equivalent for a space X, where subsequence-covering can be replaced by pseudo-sequence-covering. (1) X has a countable cs -network (resp., cs-network, sn-network).

116 110 Nguyen Van Dung No.3 (2) X is a subsequence-covering (resp., sequence-covering, 1-sequence-covering) image of a separable metric space. Recently, Y. Ge [8] sharpened Theorem 1.2 as follows. Theorem 1.5. ([8], Theorem 12) The following are equivalent for a space X. (1) X is an ℵ 0 -space. (2) X is a sequence-covering, compact-covering image of a separable metric space. (3) X is a sequentially-quotient image of a separable metric space. Taking these results into account, the following question naturally arises. Question 1.6. When P is a countable network (cs-network, sn-network, so-network) for X; how nice can the mapping f and the metric domain M be taken to? Around this question, the author of [2] has shown that, when P is a countable cs-network, f and M in Theorem 1.5 can be an mssc-mapping and a relatively compact metric space, respectively. More precisely, the following has been proved. Theorem 1.7. ([2], Theorem 2.5] The following are equivalent for a space X. (1) X is an ℵ 0 -space. (2) X is a sequence-covering, compact-covering mssc-image of a relatively compact metric space. (3) X is a sequentially-quotient image of a separable metric space. In the above result, f and M can not be any compact mapping and any compact metric space, respectively; see [2, Example 2.11 & Example 2.12]; and the key of the proof is to construct a sequence-covering, compact-covering mssc-mapping f from a relatively compact metric space M onto X. By the above, we are interested in the following question. Question 1.8. Find a consistent method to construct an mssc-mapping with coveringproperties onto a space from some relatively compact metric space? In this paper, we answer Question 1.8 by introducing the notion of σ-finite Ponomarevsystem (f, M, X, {P n }) to give a consistent method to construct an mssc-mapping f with covering-properties onto a space X from some relatively compact metric space M. As applications, we systematically obtain characterizations on mssc-images of relatively compact metric spaces under certain covering-mappings, which gives an answer for Question 1.6, particularly, for Question 1.3. Throughout this paper, all spaces are regular and T 1, N denotes the set of all natural numbers, ω = N {0}, and a convergent sequence includes its limit point. Let P be a family of subsets of X. Then P, and P denote the union {P : P P}, and the intersection {P : P P}, respectively. A sequence {xn : n ω} converging to x 0 is eventually in A X, if {x n : n n 0 } {x 0 } A for some n 0 N, and it is frequently in A if {x nk : k N} {x 0 } A for some subsequence {x nk : k N} of {x n : n N}. Definition 1.9. Let P be a family of subsets of a space X. (1) For each x X, P is a network at x in X, if x P, and if x U with U open in X, then there exists P P such that x P U. (2) P is a cs-network for X [10] if, for every convergent sequence S converging to x U with U open in X, there exists P P such that S is eventually in P U.

117 Vol.4 Notes on mssc-images of relatively compact metric spaces 111 (3) P is a cs -network for X [5] if, for every convergent sequence S converging to x U with U open in X, there exists P P such that S is frequently in P U. (4) P is a k-network of X [19] if, for every compact subset K U with U open in X, there exists a finite F P such that K F U. Definition Let P = {P x : x X} be a family of subsets of a space X satisfying that, for each x X, P x is a network at x in X, and if U, V P x, then W U V for some W P x. (1) P is a weak base for X [21], if G X such that for each x G, there exists P P x satisfying P G, then G is open in X. (2) P is an sn-network for X [14], if each member of P x is a sequential neighborhood of x in X. (3) P is an so-network for X [12], if each member of P x is sequentially open in X. (4) The above P x is respectively a weak base, an sn-network, and an so-network at x in X [12]. Definition ([4]) Let P be a subset of a space X. (1) P is a sequential neighborhood of x if, for every convergent sequence S converging to x in X, S is eventually in P. (2) P is a sequentially open subset of X if, for every x P, P is a sequential neighborhood of x. Definition Let X be a space. (1) X is relatively compact, if X is compact. (2) X is an ℵ 0 -space [18] (resp., sn-second countable space [7], so-second countable space, g- second countable space [21], second countable space [3]), if X has a countable cs-network (resp., countable sn-network, countable so-network, countable weak base, countable base). (3) X is a sequential space [4], if every sequentially open subset of X is open. Remark ([15]) (1) For a space, weak base sn-network cs-network. (2) An sn-network for a sequential space is a weak base. Remark (1) It is easy to see that compact metric relatively compact metric separable metric, and these implications can not be reversed from Example 2.14 and Example (2) It is well-known that a space X is an ℵ 0 -space if and only if X has a countable k-network (cs -network), see [23, Proposition C], for example. Remark Let f : X Y be a mapping. (1) f is an mssc-mapping[13], if X is a subspace of the product space n N X n of a family {X n : n N} of metric spaces, and for every y Y, there exists a sequence {V y,n : n N} of open neighborhoods of y in Y such that each p n (f 1 (V y,n )) is a compact subset of X n, where p n : i N X i X n is the projection. (2) f is an 1-sequence-covering mapping [14] if, for every y Y, there exists x y f 1 (y) such that whenever {y n : n N} is a sequence converging to y in Y there exists a sequence {x n : n N} converging to x y in X with each x n f 1 (y n ). (3) f is a 2-sequence-covering mapping [14] if, for every y Y, x f 1 (y), and sequence {y n : n N} converging to y in Y, there exists a sequence {x n : n N} converging to x in X

118 112 Nguyen Van Dung No.3 with each x n f 1 (y n ). (4) f is an 1-sequentially quotient mapping [17] if, for every y Y, there exists x y f 1 (y) such that whenever {y n : n N} is a sequence converging to y in Y there exists a sequence {x k : k N} converging to x y in X with each x k f 1 (y nk ). (5) f is a sequence-covering mapping [20] if, for every convergent sequence S in Y, there exists a convergent sequence L in X such that f(l) = S. (6) f is a compact-covering mapping [18] if, for every compact subset K of Y, there exists a compact subset L of X such that f(l) = K. (7) f is a pseudo-sequence-covering mapping [11], if for every convergent sequence S of Y, there exists a compact subset K of X such that f(k) = S. (8) f is a sequentially-quotient mapping [1] if, for every convergent sequence S in Y, there exists a convergent sequence L in X such that f(l) is a subsequence of S. For terms are not defined here, please refer to [3, 24]. 2. Results Definition 2.1. Let P = {P n : n N} be a countable network for a space X. Because X is T 1 and regular, we may assume that every member of P is closed. For each n N, let P n = {X} {P i : i n} = {P α : α A n }, where A n is a finite set, and let every A n be endowed with the discrete topology. Setting M = { a = (α i ) } A i : {P αi : i N} forms a network at some point x a X. i N Then M, which is a subspace of the product space i N A i, is a metric space. Since X is T 1 and regular, x a is unique for each a M. We define f : M X by f(a) = x a for each a M. Then f is an mssc-mapping from a relatively compact metric space onto X by the following Theorem 2.3. The system (f, M, X, {P n }) is a σ-finite Ponomarev-system. Remark 2.2. For more details on Ponomarev-systems and images of metric spaces, see [9,16,25], for example. Theorem 2.3. Let (f, M, X, {P n }) be the system in Definition 2.1. Then the following hold. (1) M is a relatively compact metric space. (2) f is an mssc-mapping. Proof. (1) For every i N, since A i is finite, A i is compact metric. Then i N A i is compact metric, and so M i N A i is. It implies that M is relatively compact metric. (2) It suffices to prove the following facts (a), (b), and (c). (a) f is onto. Let x X. Since P is a countable network for X, (P) x = {P P : x P } is a countable network at x in X. We may assume that (P) x = {P x,j : j N}, where P x,j P i(j) for some

119 Vol.4 Notes on mssc-images of relatively compact metric spaces 113 i(j) N satisfying i(j) < i(j + 1). For each i N, taking P αi = P i(j) if i = i(j) for some j N, and otherwise, P αi = X. Then (α i ) i N A i, and {P αi : i N} = (P) x forms a network at x in X. Setting a = (α i ), we get a M and f(a) = x. (b) f is continuous. Let x = f(a) U with U open in X and a M. Setting a = (α i ) i N A i, where {P αi : i N} forms a network at x in X. Then there exists n N such that x P αn U. Setting M a = {b = (β i ) M : β n = α n }, then M a is an open neighborhood of a in M. For each b M a, we get f(b) P βn = P αn U. It implies that f(m a ) U. (c) f is an mssc-mapping. Let x X. For each i N, taking V x,i = X. Then {V x,i : i N} is a sequence of open neighborhoods of x in X. Since A i is finite, A i is compact. Then p i (f 1 (V x,i )) = p i (f 1 (X)) A i is compact. It implies that f is an mssc-mapping. Remark 2.4. In the view of (a) in the proof (1) (2) of [2, Theorem 2.5], f is onto by choosing a = (α i ) satisfying f(a) = x, where x P αi P i. Unfortunately, this argument is not true. In fact, we may pick x P αi = X P i. Then {P αi : i N} = {X} does not form a network at x in X. It is a contradiction. The wrong argument is corrected as in the proof (2).(a) of Theorem 2.3. Theorem 2.5. Let (f, M, X, {P n }) be a σ-finite Ponomarev-system. Then the following hold. (1) If P is a cs-network for X, then f is sequence-covering. (2) If P is a k-network for X, then f is a compact-covering. (3) If P is a cs -network for X, then f is pseudo-sequence-covering. (4) If P is an sn-network for X, then f is 1-sequence-covering. (5) If P is an so-network for X, then f is 2-sequence-covering. Proof. (1) Let S = {x m : m ω} be a convergent sequence converging to x 0 in X. Suppose that U is an open neighborhood of S in X. A family A of subsets of X has property cs(s, U) if: (i) A is finite. (ii) for each P A, = P S P U. (iii) for each x m S, there exists unique P xm A such that x m P xm. (iv) if x 0 P A, then S P is finite. For each i N, since A = {X} P i has property cs(s, X) and P i is finite, we can assume that {A P i : A has property cs(s, X)} = {A i(j) : j = n i 1 + 1,..., n i }, where n 0 = 0. By this notation, for each j N, there is unique i(j) N such that A i(j) has property cs(s, X). Then for each j N, we can put A i(j) = {P α : α E j }, where E j is a finite subset of A j. For each j N, m ω and x m S, it follows from (iii) that there is unique α j,m E j such that x m P αj,m A i(j). Let a m = (α j,m ) E j A j. We shall prove that j N j N {P αj,m : j N} is a network at x m in X. In fact, let x m U with U open in X, we consider

120 114 Nguyen Van Dung No.3 two following cases. (a) If m = 0, then S is eventually in P x0 U for some P x0 P. For each x S P x0, let x P x X (S {x}) for some P x P. Then G = {P x0 } {P x : x S P x0 } P has property cs(s, X). Since G is finite, G P i for some i N. It implies that G = A i(j) for some i(j) N with j {n i 1 + 1,..., n i }. Since x m = x 0 P αj,0, P αj,0 = P x0. Hence x m = x 0 P αj,0 U. (b) If m 0, then S {x m } is eventually in P x0 X {x m } for some P x0 P. For each x (S {x m }) P x0, let x P x X (S {x}) for some P x P, and let x m P xm U (X (S {x m })) for some P xm P. Then H = {P x0 } {P xm } {P x : x (S {x m }) P x0 } has property cs(s, X). Since H is finite, H P i for some i N. It implies that H = A i(j) for some i(j) N with j {n i 1 + 1,..., n i }. Since x m P αj,m, P αj,m = P xm. Hence x m P αj,m U. By the above, for each m ω we get a m = (α j,m ) M satisfying f(a m ) = x m. For each j N, since families H and G are finite, there exists m j N such that α j,m = α j,0 if m m j. Hence the sequence {α j,m : m N} converges to α j,0 in A j. Thus, the sequence {a m : m N} converges to a 0 in M. Setting L = {a m : m ω}, then L is a convergent sequence in M and f(l) = S. This shows that f is sequence-covering. (2)Let K be a compact subset of X. Suppose that V is an open neighborhood of K in X. A family B of subsets of X has property k(k, V ) if: (i) B is finite. (ii) P K for each P B. (iii) K B V. For each i N, since B = {X} P i has property k(k, X) and P i is finite, we can assume that {B P i : B has property k(k, X)} = {B i(j) : j = n i 1 + 1,..., n i }, where n 0 = 0. By this notation, for each j N, there is unique i(j) N such that B i(j) has property k(k, X). Then for each j N, we can put B i(j) = {P α : α F j }, where F j is a finite subset of A j. Setting L = {a = (α i ) F i : P αi ) }, we shall prove that L is a compact i N i N(K subset of M satisfying f(l) = K, hence f is compact-covering, by the following facts (a), (b), and (c). (a) L is compact. Since L F i and F i is compact, we only need to prove that L is closed in F i. Let i N i N i N a = (α i ) F i L. Then P αi ) =. Since K P αi is compact, there exists i 0 N such i N i N(K that (K P αi ) =. Setting W = {b = (β i ) F i : β i = α i if i i 0 }. Then W is an open i i 0 i N neighborhood of a in F i and W L =. If not, there exists b = (β i ) W L. Since b L, i N (K P βi ), hence i N i i 0 (K P βi ). Since b W, i i 0 (K P αi ) = i i 0 (K P βi ).

121 Vol.4 Notes on mssc-images of relatively compact metric spaces 115 This is a contradiction of the fact that (K P αi ) =. i i 0 (b) L M and f(l) K. Let a = (α i ) L, then a F i and (K P αi ). Taking x P αi ). If i N i N i N(K {P αi : i N} is a network at x in X, then a M and f(a) = x, hence L M and f(l) K. So we only need to prove that {P αi : i N} is a network at x in X. Let V be an open neighborhood of x in X. There exist an open subset W of K such that x W, and compact subsets W K and K W such that W K V and K W X {x}, where W K is the closure of W in K. Since P is a k-network for X, there exist finite families F 1 P and F 2 P such that W K F 1 V and K W F 2 X {x}. We may assume that P K for each P F 1 F 2. Setting F = F 1 F 2, then F has property k(k, X). It implies that F = B i(j) for some i(j) N with j {n i 1 + 1,..., n i }. Since x P αj B i(j), P αj F 1. Then x P αj V. This prove that {P αj : j N} is a network at x in X. (c) K f(l). Let x K. For each i N, there exists α i F i such that x P αi. Setting a = (α i ), then a L. Furthermore, f(a) = x as in the proof of (b). So K f(l). (3) Let S be a convergent sequence of X. By using [26, Lemma 3] and as in the proof of (2), here S plays the role of the compact subset K, there exists a compact subset L of M such that f(l) = S. Then, f is pseudo-sequence-covering. (4) Let P = {P x : x X}, where each P x is an sn-network at x in X. Let x X. We may assume that P x = {P x,j : j N}, where P x,j P i(j) with some i(j) N satisfying i(j) < i(j+1). For each i N, take P αi = P i(j) if i = i(j) for some j N, and otherwise, P αi = X. Then (α i ) A i, and {P αi : i N} = P x forms a network at x in X. Setting a x = (α i ), then i N a M and a x f 1 (x). For each n N, setting B n = {b = (β i ) M : β i = α i if i n}. Then {B n : n N} is a decreasing neighborhood base at a x in M. For each n N, we get f(b n ) = P βi P βi = P αi if i n. Then P αi. In fact, for each b = (β i ) B n, f(b) = i n i N f(b) P αi. It implies that f(b n ) P αi. Conversely, if x P αi, then x = f(b) for i n i n some b = (β i ) M. Setting c = (γ i ), where γ i = α i if i n, and γ i = β i n if i > n. Then {P γi : i N} is a network at x in X. For each i n, since β i n A i n A i, γ i A i. Then c B n and f(c) = x. It implies that P αi f(b n ). i n Suppose that the sequence {x j : j N} converges to x in X. For each i N, since P αi is a sequential neighborhood of x in X, there exists j i N such that x j P αi for each j j i. For each n N, setting j(n) = max{j i : i n}, then x j = f(b n ) for each j j(n). It implies that f 1 (x j ) B n for each j j(n). We may assume that 1 < j(n) < j(n + 1). For each j N, taking f 1 (x j ) if j < j(1), a j f 1 (x j ) B n if j(n) j < j(n + 1). Then the sequence {a j : j N} converges to a x in M. In fact, let U be a neighborhood of i n P αi i n

122 116 Nguyen Van Dung No.3 a x, then there exists n N such that a x B n U. Obviously, a j B n U for each j j(n), so {a j : j N} converges to a x. Note that f(a j ) = x j for every j N. It implies that f is 1-sequence-covering. (5) Let x X and a x f 1 (x). Setting a x = (α i ), then α i A i for each i N, and {P αi : i N} forms a network at x in X. For each n N, setting B n = {b = (β i ) M : β i = α i if i n}. Then {B n : n N} is a decreasing neighborhood base at a x in M, and for each n N, f(b n ) = i n P α i as in the proof of (4). Suppose that sequence {x j : j N} converges to x in X. For each i N, since P αi is sequentially open, there exists j i N such that x j P αi for each j j i. As in the proof (4), we get a sequence {a j : j N} converges to a x in M such that f(a j ) = x j for each j N. It implies that f is 2-sequence-covering. By using Theorem 2.3 and Theorem 2.5 above, we obtain characterizations of mssc-images of relatively compact metric spaces under certain covering-mappings, which sharpen results in [18,26] and more. Firstly, we sharpen Theorem 1.1 as following. Corollary 2.6. The following are equivalent for a space X. (1) X is a cosmic space. (2) X is an mssc-image of a relatively compact metric space. (3) X is an image of a separable metric space. Proof. We only need to prove (1) (2), other implications are routine. Since X is a cosmic space, X has a countable network P. Then the σ-finite Ponomarev-system (f, M, X, {P n }) exists. Therefore, f is an mssc-mapping and M is relatively compact metric by Theorem 2.3. It implies that X is an mssc-image of a relatively compact metric space. Remark 2.7. Corollary 2.6 is an answer of Question 1.3. Next, we extend partly Theorem 1.4 as follows. Corollary 2.8. The following are equivalent for a space X. (1) X is an sn-second countable space. (2) X is an 1-sequence-covering, compact-covering mssc-image of a relatively compact metric space. (3) X is an 1-sequentially-quotient image of a separable metric space. Proof. (1) (2). Since X is an sn-second countable space, X has a countable closed sn-network P 1 and a countable closed k-network P 2. Set P = P 1 P 2, then P is a countable snnetwork and k-network for X. It follows from Definition 2.1 that the σ-finite Ponomarev-system (f, M, X, {P n }) exists. Therefore, f is an 1-sequence-covering, compact-covering mssc-mapping from a relatively compact metric space M onto X by Theorem 2.3 and Theorem 2.5. (2) (3). It is obvious. (3) (1). Let f : M X be an 1-sequentially-quotient mapping from a separable metric space M onto X. Since M is separable metric, M has a countable base B. We may assume that B is closed under finite intersections. For each x X, there exists a x f 1 (x) such that whenever {x n : n N} is a sequence converging to x in X there exists a sequence {a k : k N} converging to a x in M with each a k f 1 (x nk ). Let B x = {B B : a x B} and P x = f(b x ). We shall prove that P = {P x : x X} is a countable sn-network for X. Since B is countable, P is countable. It suffices to prove the following facts (a), (b), and (c) for every

123 Vol.4 Notes on mssc-images of relatively compact metric spaces 117 x X. (a) P x is a network at x in X. It is straightforward because B x is a neighborhood base at a x in M. (b) If P 1, P 2 P x, then P P 1 P 2 for some P P x. Let P 1 = f(b 1 ), P 2 = f(b 2 ) with B 1, B 2 B x. Then B 1 B 2 B x. Setting P = f(b 1 B 2 ), we get P P x and P P 1 P 2. (c) Each P P x is a sequential neighborhood of x. Let P = f(b) with B B x, and let {x n : n N} be a sequence converging to x in X. For each subsequence {x ni : i N} of {x n : n N}, since f is 1-sequentially-quotient, there exists a sequence {a k : k N} converging to a x in M with each a k f 1 (x ni,k ). Since {a k : k N} {a x } is eventually in B, {x ni : i N} {x} is frequently in P. Then {x n : n N} {x} is frequently in P. It follows from [6, Remark 1.4] that P is a sequential neighborhood of x. Corollary 2.9. The following are equivalent for a space X. (1) X is a g-second countable space. (2) X is an 1-sequence-covering, compact-covering, quotient mssc-image of a relatively compact metric space. (3) X is an 1-sequentially-quotient, quotient image of a separable metric space. Proof. (1) (2). It follows from Corollary 2.8 that X is an 1-sequence-covering msscimage of a relatively compact metric space under some mapping f. Since f is an 1-sequencecovering mapping onto a g-second countable space, f is a quotient mapping by [15, Lemma 3.5]. It implies that X is an 1-sequence-covering, quotient mssc-image of a relatively compact metric space. (2) (3). It is obvious. (3) (1). Since X is a quotient image of a separable metric space, X is a sequential space by [4, Proposition 1.2]. Then X is a sequential space having a countable sn-network P by Corollary 2.8. It follows from Lemma 1.13 that P is a weak base for X. Then X is a g-second countable space. In the following, we prove a result on spaces having countable so-networks. Corollary The following are equivalent for a space X. (1) X is an so-second countable space. (2) X is a 2-sequence-covering, compact-covering mssc-image of a relatively compact metric space. (3) X is a 2-sequence-covering image of a separable metric space. Proof. (1) (2). Since X is an so-second countable space, X has a countable closed so-network P 1 and a countable closed k-network P 2. Set P = P 1 P 2, then P is a countable sonetwork and k-network for X. It follows from Definition 2.1 that the σ-finite Ponomarev-system (f, M, X, {P n }) exists. Therefore, f is a 2-sequence-covering, compact-covering mssc-mapping from a relatively compact metric space M onto X by Theorem 2.3 and Theorem 2.5. (2) (3). It is obvious. (3) (1). Let f : M X be a 2-sequence-covering mapping from a separable metric space M onto X. Since M is separable metric, M has a countable base B. For each x X,

124 118 Nguyen Van Dung No.3 let B x = {B B : f 1 (x) B }, and P x be the family of finite intersections of members of f(b x ). We shall prove that P = {P x : x X} is a countable so-network for X. Since B is countable, P is countable. It suffices to prove the following facts (a), (b), and (c) for every x X. (a) P x is a network at x in X. It is straightforward because B x is a base of f 1 (x) in M. (b) If P 1, P 2 P x, then P P 1 P 2 for some P P x. It is straightforward by choosing P = P 1 P 2. (c) Each P P x is sequentially open. For each y f(b) with B B x, we get f 1 (y) B. Let {y n : n N} be a sequence converging to y in X. Then there exists a sequence {a n : n N} converging to a f 1 (y) B such that f(a n ) = y n for each n N. Since B is open, {a n : n N} {a} is eventually in B. It implies that {y n : n N} {y} is eventually in f(b). Hence f(b) is a sequential neighborhood of y in X. Therefore, f(b) is sequentially open. Since each P P x is some intersection of finitely many members of f(b x ), P is sequentially open. Corollary The following are equivalent for a space X. (1) X is a second countable space. (2) X is a 2-sequence-covering, compact-covering, quotient mssc-image of a relatively compact metric space. (3) X is a 2-sequence-covering, quotient image of a separable metric space. Proof. (1) (2). It follows from Corollary 2.10 that X is a 2-sequence-covering msscimage of a relatively compact metric space under some mapping f. Since f is a 2-sequencecovering mapping onto a second countable space, f is a quotient mapping by [15, Lemma 3.5]. It implies that X is a 2-sequence-covering, quotient mssc-image of a relatively compact metric space. (2) (3). It is obvious. (3) (1). Since X is a quotient image of a separable metric space, X is a sequential space by [4, Proposition 1.2]. Then X is a sequential space having a countable so-network P by Corollary For each P P, since P is sequentially open and X is sequential, P is open. Then P is a countable base for X, i.e., X is a second countable space. Remark By using Theorem 2.3 and Theorem 2.5, we also get the proof of Theorem 1.7 again, which was presented in [2]. From the above results, we obtain preservations of certain spaces under covering-mappings as follows. Corollary Let f : X Y be a mapping. Then the following hold. (1) If X is an sn-second countable space and f is 1-sequentially-quotient, then Y is an sn-second countable space. (2) If X is a g-second countable space and f is 1-sequentially-quotient, quotient, then Y is a g-second countable space. (3) If X is an so-second countable space and f is 2-sequence-covering, then Y is an so-second countable space. (4) If X is a second countable space and f is 2-sequence-covering, quotient, then Y is a second

125 Vol.4 Notes on mssc-images of relatively compact metric spaces 119 countable space. Proof. We only need to prove (1), by similar arguments, we get the other. Since X is an sn-second countable space, X is an 1-sequentially-quotient image of a separable metric space under some mapping g by Corollary 2.8. It is easy to see that f g is also an 1-sequentiallyquotient mapping. Then Y is an 1-sequentially-quotient image of a separable metric space under the mapping f g. Therefore, Y is an sn-second countable space by Corollary 2.8. Finally, we give examples to illustrate the above results. Example ([2], Example 2.8) A relatively compact metric space is not compact. Example ([2], Example 2.9) A separable metric space is not relatively compact. Example A 2-sequence-covering, compact-covering mapping from a separable metric space is not an mssc-mapping. Proof. Let f be the mapping in [2, Example 2.10]. Then f is also 2-sequence-covering. This complete the proof. Example A second countable space is not any image of a compact metric space. It implies that relatively compact metric in the above results can not be replaced by compact metric. Proof. Let R be the set of all real numbers endowed with the usual topology. Then R is a second countable space. Since R is not compact, R is not any image of a compact metric space. Example A g-second countable space is not any 1-sequence-covering, compactcovering compact image of a separable metric space. It implies that mssc-image in Corollary 2.8 can not be replaced by compact image. Proof. See [22, Example 2.14(3)]. References [1] J. R. Boone and F. Siwiec, Sequentially quotient mappings, Czechoslovak Math. J., 26 (1976), [2] N. V. Dung, ℵ 0 -spaces and mssc-images of relatively compact metric spaces, Sci. Magna, 4(2008), No.2, [3] R. Engelking, General topology, Sigma series in pure mathematics, Heldermann Verlag, Berlin, 6(1988). [4] S. P. Franklin, Spaces in which sequences suffice, Fund. Math., 57(1965), [5] Z. M. Gao, ℵ-space is invariant under perfect mappings, Questions Answers General Top., 5(1987), [6] Y. Ge, On compact images of locally separable metric spaces, Topology Proc., 27(2003), No.1, [7] Y. Ge, Spaces with countable sn-networks, Comment. Math. Univ. Carolin, 45 (2004), [8] Y. Ge, ℵ 0 -spaces and images of separable metric sapces, Siberian Elec. Math. Rep., 74 (2005), [9] Y. Ge, Mappings in Ponomarev-systems, Topology Proc., 29 (2005), No.1,

126 120 Nguyen Van Dung No. 3 [12] Z. Li, Q. Li, and X. Zhou, On sequence-covering msss-maps, Mat. Vesnik, 59 (2007), [13] S. Lin, Locally countable families, locally finite families and Alexandroff problem, Acta Math. Sinica (Chin. Ser.), 37 (1994), [14] S. Lin, On sequence-covering s-mappings, Adv. Math (China), 25 (1996), [15] S. Lin and P. Yan, Sequence-covering maps of metric spaces, Topology Appl., 109 (2001), [16] S. Lin and P. Yan, Notes on cfp-covers, Comment. Math. Univ. Carolin, 44 (2003), No.2, [17] S. Lin, J-C. Zhu, Y. Ge, and J-S. Gu, Almost-open maps, sequence-covering maps and sn-networks, Indian J. Pure Appl. Math., 37 (2006), No.2, [18] E. Michael, ℵ 0 -spaces, J. Math. Mech., 15(1966), [19] P. O Meara, On paracompactness in function spaces with the compact-open topology, Proc. Amer. Math. Soc, 29 (1971), [20] F. Siwiec, Sequence-covering and countably bi-quotient mappings, General Topology Appl., 1 (1971), [21] F. Siwiec, On defining a space by a weak-base, Pacific J. Math, 52 (1974), [22] Y. Tanaka, Symmetric spaces, g-developable spaces and g-metrizable spaces, Math. Japonica, 36 (1991), [23] Y. Tanaka, Theory of k-networks, Questions Answers in Gen, Topology, 12 (1994), [24] Y. Tanaka, Theory of k-networks II, Questions Answers in Gen, Topology, 19 (2001), [25] Y. Tanaka and Y. Ge, Around quotient compact images of metric spaces, and symmetric spaces, Houston J. Math., 32 (2006), No.1, [26] Y. Tanaka and Z. Li, Certain covering-maps and k-networks, and related matters, Topology Proc., 27 (2003), No.1,

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